Problems In each of Problems 1 through 11: a. Seek power series solutions of the given differential equation about the given point xo: find the recurrence relation that the coefficients must satisfy b. Find the first four nonzero terms in each of two solutions y and 17. Show directly, using the ratio test, that the two series s of Airy's equation about x 0 converge for all x; see equa of the text. 18. The Hermite Equation. The equation (unless the series terminates sooner). y"-2xy'+Ay 0, Y c. By evaluating the Wronskian Wlyi, y2l(o), show that y and -000 34o0, where A is a constant, is known as the Hermite equatio important equation in mathematical physics. a. Find the first four nonzero terms in each of twe about x form a fundamental set of solutions. yS d. If possible, find the general term in each solution. 1. y-y= 0, o= 0 2. y"+3y 0, xo0 3. y-xy-y = 0, xo 0 4. y"-xy'-y = 0, 5. y"+k2xy 0, xo = 6. xo 0 0 and show that they form a fundame solutions. b. Observe that if A is a nonnegative even intege or the other of the series solutions terminates and Xo = 1 polynomial. Find the polynomial solutions for X 8, and 10. Note that each polynomial is determined multiplicative constant. c. The Hermite polynomial H (x) is defined as th solution of the Hermite equation with A = 2n f coefficient of x" is 2". Find Ho(x), H(x), .. E 0, k a constant (1-x)y"+y = 0, 7. = 0 y" +xy'+2y 0, 8. xy"+y'+xry 0, (3-x)y"- 3xy' - y= 0, Xo= 0 xo 1 xo 0 9. 19. Consider the initial-value problem y' = v1- y a. Show that y = sinx is the solution of thi problem. b. Look for a solution of the initial-value problem Xo 0 0, 2y" +xy' +3y 10. 2 xo 2y" + (x + 1) y' + 3y = 0, 11. In each of Problems 12 through 14: 0. Find the coefficients a nower series about x = firt five nonzero terms in the solution of the given ewl+53'13 2.3-3 . For this equation P(x) = 1. Q0x) = 0, and R(x) =-x hence every point is an ordinary point. We assume that dy(x) + ax) (16) where y(x) and ys(x) are the first and second bracketed expre Having obtained these two series solutions, we can now inv of the rapid growth of the denominators of the terms in the serie expect these series to have a large radius of convergence. Inde show that both of these series converge for all x; see Problem Assume for the moment that the series for y and y do c first a 1, a y and that the series converges in some interval lx!< p.The series for y" is given by equation (7); as explained in the preceding example, we can rewrite it as (17) (n+2)(n+1)a,+2 = 0 and then do = 0, a ot equation (15). Notice that y satisfies the initial condition satisfies the initial conditions y2(0)-0,y,(0) = 1. Thus W y and yy are a fundamental set of solutions, Hence the gener 1, it follows thar NO Substituting the series (16) and (17) for y and y" into the left-hand side of equation (15), we obtain y@(x) + a(x) (n+2)(+ Da (18) ax=(n+2)(n+ 1)a,.x-ax -00 In Figures 5.2.3 and 5.2.4, respectively, we show the gra equation as well as graphs of several partial sums of the t *e Next, we shift the index of summation in the second series on the right-hand side of equation (s) by replacing n byn -1 and starting the summation at 1 rather than zero. Thus we write equation (15) partial sums provide local approximations to the solutions in the quality of the approximation improves as the number adequately represent yi and yy for large Jx]. A practical given partial sum is reasonably accurate is to compare the one, obtained by including one more term. As soon as the can be confident that the original partial sum is no longer as a"-0. 2- lag+n+2)(n+ Dx- l graphs for n= *-l 24 and n 27 begin to separate at about partial sum of degree 24 is worthless as an approximation Again, for this equation to be satisfied for allx in some interval, the coefficients of like powers of x must be zero; hence a = 0, and we obtain the recurrence relation X (19) (n+2)(n+1)a2-a-10 for n 1,2,3, .... 24 36 n 48 12 Since a is given in terms of a,.t, the a's are determined in steps of three. Thus ao determines a3, which in turn determines as,..a, determines a4, which in turn determines a,. and ay determines as, which in turn determines as, . .Since az = 0, we immediately conclude that as = as = a1= 0. For the sequence ao, as, as, ag, we set n= 1, 4, 7, 10, ... in the recurrence relation: 18 61 42 30 y ya) AA do -8 a3 06 8-9 2-3-5 6-8-9 2-3-5.6 5 6 2-3 These results suggest the general formula do n 4. 2.3-5-6(3n - 1)(3n) o
Problems In each of Problems 1 through 11: a. Seek power series solutions of the given differential equation about the given point xo: find the recurrence relation that the coefficients must satisfy b. Find the first four nonzero terms in each of two solutions y and 17. Show directly, using the ratio test, that the two series s of Airy's equation about x 0 converge for all x; see equa of the text. 18. The Hermite Equation. The equation (unless the series terminates sooner). y"-2xy'+Ay 0, Y c. By evaluating the Wronskian Wlyi, y2l(o), show that y and -000 34o0, where A is a constant, is known as the Hermite equatio important equation in mathematical physics. a. Find the first four nonzero terms in each of twe about x form a fundamental set of solutions. yS d. If possible, find the general term in each solution. 1. y-y= 0, o= 0 2. y"+3y 0, xo0 3. y-xy-y = 0, xo 0 4. y"-xy'-y = 0, 5. y"+k2xy 0, xo = 6. xo 0 0 and show that they form a fundame solutions. b. Observe that if A is a nonnegative even intege or the other of the series solutions terminates and Xo = 1 polynomial. Find the polynomial solutions for X 8, and 10. Note that each polynomial is determined multiplicative constant. c. The Hermite polynomial H (x) is defined as th solution of the Hermite equation with A = 2n f coefficient of x" is 2". Find Ho(x), H(x), .. E 0, k a constant (1-x)y"+y = 0, 7. = 0 y" +xy'+2y 0, 8. xy"+y'+xry 0, (3-x)y"- 3xy' - y= 0, Xo= 0 xo 1 xo 0 9. 19. Consider the initial-value problem y' = v1- y a. Show that y = sinx is the solution of thi problem. b. Look for a solution of the initial-value problem Xo 0 0, 2y" +xy' +3y 10. 2 xo 2y" + (x + 1) y' + 3y = 0, 11. In each of Problems 12 through 14: 0. Find the coefficients a nower series about x = firt five nonzero terms in the solution of the given ewl+53'13 2.3-3 . For this equation P(x) = 1. Q0x) = 0, and R(x) =-x hence every point is an ordinary point. We assume that dy(x) + ax) (16) where y(x) and ys(x) are the first and second bracketed expre Having obtained these two series solutions, we can now inv of the rapid growth of the denominators of the terms in the serie expect these series to have a large radius of convergence. Inde show that both of these series converge for all x; see Problem Assume for the moment that the series for y and y do c first a 1, a y and that the series converges in some interval lx!< p.The series for y" is given by equation (7); as explained in the preceding example, we can rewrite it as (17) (n+2)(n+1)a,+2 = 0 and then do = 0, a ot equation (15). Notice that y satisfies the initial condition satisfies the initial conditions y2(0)-0,y,(0) = 1. Thus W y and yy are a fundamental set of solutions, Hence the gener 1, it follows thar NO Substituting the series (16) and (17) for y and y" into the left-hand side of equation (15), we obtain y@(x) + a(x) (n+2)(+ Da (18) ax=(n+2)(n+ 1)a,.x-ax -00 In Figures 5.2.3 and 5.2.4, respectively, we show the gra equation as well as graphs of several partial sums of the t *e Next, we shift the index of summation in the second series on the right-hand side of equation (s) by replacing n byn -1 and starting the summation at 1 rather than zero. Thus we write equation (15) partial sums provide local approximations to the solutions in the quality of the approximation improves as the number adequately represent yi and yy for large Jx]. A practical given partial sum is reasonably accurate is to compare the one, obtained by including one more term. As soon as the can be confident that the original partial sum is no longer as a"-0. 2- lag+n+2)(n+ Dx- l graphs for n= *-l 24 and n 27 begin to separate at about partial sum of degree 24 is worthless as an approximation Again, for this equation to be satisfied for allx in some interval, the coefficients of like powers of x must be zero; hence a = 0, and we obtain the recurrence relation X (19) (n+2)(n+1)a2-a-10 for n 1,2,3, .... 24 36 n 48 12 Since a is given in terms of a,.t, the a's are determined in steps of three. Thus ao determines a3, which in turn determines as,..a, determines a4, which in turn determines a,. and ay determines as, which in turn determines as, . .Since az = 0, we immediately conclude that as = as = a1= 0. For the sequence ao, as, as, ag, we set n= 1, 4, 7, 10, ... in the recurrence relation: 18 61 42 30 y ya) AA do -8 a3 06 8-9 2-3-5 6-8-9 2-3-5.6 5 6 2-3 These results suggest the general formula do n 4. 2.3-5-6(3n - 1)(3n) o
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
How do I complete problem #7 from part A to D? This is from my
Transcribed Image Text:Problems
In each of Problems 1 through 11:
a. Seek power series solutions of the given differential equation
about the given point xo: find the recurrence relation that the
coefficients must satisfy
b. Find the first four nonzero terms in each of two solutions y
and
17. Show directly, using the ratio test, that the two series s
of Airy's equation about x 0 converge for all x; see equa
of the text.
18. The Hermite Equation. The equation
(unless the series terminates sooner).
y"-2xy'+Ay 0,
Y
c. By evaluating the Wronskian Wlyi, y2l(o), show that y
and
-000 34o0,
where A is a constant, is known as the Hermite equatio
important equation in mathematical physics.
a. Find the first four nonzero terms in each of twe
about x
form a fundamental set of solutions.
yS
d. If possible, find the general term in each solution.
1. y-y= 0, o= 0
2. y"+3y 0, xo0
3. y-xy-y = 0, xo 0
4. y"-xy'-y = 0,
5. y"+k2xy 0, xo =
6. xo 0
0 and show that they form a fundame
solutions.
b. Observe that if A is a nonnegative even intege
or the other of the series solutions terminates and
Xo = 1
polynomial. Find the polynomial solutions for X
8, and 10. Note that each polynomial is determined
multiplicative constant.
c. The Hermite polynomial H (x) is defined as th
solution of the Hermite equation with A = 2n f
coefficient of x" is 2". Find Ho(x), H(x), .. E
0, k a constant
(1-x)y"+y = 0,
7.
= 0
y" +xy'+2y 0,
8. xy"+y'+xry 0,
(3-x)y"- 3xy' - y= 0,
Xo= 0
xo 1
xo 0
9.
19. Consider the initial-value problem y' = v1- y
a. Show that y = sinx is the solution of thi
problem.
b. Look for a solution of the initial-value problem
Xo 0
0,
2y" +xy' +3y
10.
2
xo
2y" + (x + 1) y' + 3y = 0,
11.
In each of Problems 12 through 14:
0. Find the coefficients
a nower series about x =
firt five nonzero terms in the solution of the given
![ewl+53'13
2.3-3
.
For this equation P(x) = 1. Q0x) = 0, and R(x) =-x hence every point is an ordinary point.
We assume that
dy(x) + ax)
(16)
where y(x) and ys(x) are the first and second bracketed expre
Having obtained these two series solutions, we can now inv
of the rapid growth of the denominators of the terms in the serie
expect these series to have a large radius of convergence. Inde
show that both of these series converge for all x; see Problem
Assume for the moment that the series for y and y do c
first a 1, a
y
and that the series converges in some interval lx!< p.The series for y" is given by equation (7); as
explained in the preceding example, we can rewrite it as
(17)
(n+2)(n+1)a,+2
= 0 and then do = 0, a
ot equation (15). Notice that y satisfies the initial condition
satisfies the initial conditions y2(0)-0,y,(0) = 1. Thus W
y and yy are a fundamental set of solutions, Hence the gener
1, it follows thar
NO
Substituting the series (16) and (17) for y and y" into the left-hand side of equation (15), we obtain
y@(x) + a(x)
(n+2)(+ Da
(18)
ax=(n+2)(n+ 1)a,.x-ax
-00
In Figures 5.2.3 and 5.2.4, respectively, we show the
gra
equation as well as graphs of several partial sums of the t
*e
Next, we shift the index of summation in the second series on the right-hand side of equation (s) by
replacing n byn -1 and starting the summation at 1 rather than zero. Thus we write equation (15)
partial sums provide local approximations to the solutions in
the quality of the approximation improves as the number
adequately represent yi and yy for large Jx]. A practical
given partial sum is reasonably accurate is to compare the
one, obtained by including one more term. As soon as the
can be confident that the original partial sum is no longer
as
a"-0.
2- lag+n+2)(n+ Dx-
l
graphs for n=
*-l
24 and n 27 begin to separate at about
partial sum of degree 24 is worthless as an approximation
Again, for this equation to be satisfied for allx in some interval, the coefficients of like powers of x
must be zero; hence a = 0, and we obtain the recurrence relation
X
(19)
(n+2)(n+1)a2-a-10 for n 1,2,3, ....
24
36
n 48
12
Since a is given in terms of a,.t, the a's are determined in steps of three. Thus ao determines
a3, which in turn determines as,..a, determines a4, which in turn determines a,. and ay
determines as, which in turn determines as, . .Since az = 0, we immediately conclude that
as = as = a1= 0.
For the sequence ao, as, as, ag, we set n= 1, 4, 7, 10, ... in the recurrence relation:
18
61
42 30
y ya)
AA
do
-8
a3
06
8-9 2-3-5 6-8-9
2-3-5.6
5 6
2-3
These results suggest the general formula
do
n 4.
2.3-5-6(3n - 1)(3n)
o](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F81ed2a7d-ef58-4ac0-a29f-af1e2992bb8e%2F7b6cec1f-a732-4d20-9241-661ae3c12296%2Ftxyyplo.jpeg&w=3840&q=75)
Transcribed Image Text:ewl+53'13
2.3-3
.
For this equation P(x) = 1. Q0x) = 0, and R(x) =-x hence every point is an ordinary point.
We assume that
dy(x) + ax)
(16)
where y(x) and ys(x) are the first and second bracketed expre
Having obtained these two series solutions, we can now inv
of the rapid growth of the denominators of the terms in the serie
expect these series to have a large radius of convergence. Inde
show that both of these series converge for all x; see Problem
Assume for the moment that the series for y and y do c
first a 1, a
y
and that the series converges in some interval lx!< p.The series for y" is given by equation (7); as
explained in the preceding example, we can rewrite it as
(17)
(n+2)(n+1)a,+2
= 0 and then do = 0, a
ot equation (15). Notice that y satisfies the initial condition
satisfies the initial conditions y2(0)-0,y,(0) = 1. Thus W
y and yy are a fundamental set of solutions, Hence the gener
1, it follows thar
NO
Substituting the series (16) and (17) for y and y" into the left-hand side of equation (15), we obtain
y@(x) + a(x)
(n+2)(+ Da
(18)
ax=(n+2)(n+ 1)a,.x-ax
-00
In Figures 5.2.3 and 5.2.4, respectively, we show the
gra
equation as well as graphs of several partial sums of the t
*e
Next, we shift the index of summation in the second series on the right-hand side of equation (s) by
replacing n byn -1 and starting the summation at 1 rather than zero. Thus we write equation (15)
partial sums provide local approximations to the solutions in
the quality of the approximation improves as the number
adequately represent yi and yy for large Jx]. A practical
given partial sum is reasonably accurate is to compare the
one, obtained by including one more term. As soon as the
can be confident that the original partial sum is no longer
as
a"-0.
2- lag+n+2)(n+ Dx-
l
graphs for n=
*-l
24 and n 27 begin to separate at about
partial sum of degree 24 is worthless as an approximation
Again, for this equation to be satisfied for allx in some interval, the coefficients of like powers of x
must be zero; hence a = 0, and we obtain the recurrence relation
X
(19)
(n+2)(n+1)a2-a-10 for n 1,2,3, ....
24
36
n 48
12
Since a is given in terms of a,.t, the a's are determined in steps of three. Thus ao determines
a3, which in turn determines as,..a, determines a4, which in turn determines a,. and ay
determines as, which in turn determines as, . .Since az = 0, we immediately conclude that
as = as = a1= 0.
For the sequence ao, as, as, ag, we set n= 1, 4, 7, 10, ... in the recurrence relation:
18
61
42 30
y ya)
AA
do
-8
a3
06
8-9 2-3-5 6-8-9
2-3-5.6
5 6
2-3
These results suggest the general formula
do
n 4.
2.3-5-6(3n - 1)(3n)
o
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