bessel- legendre de 1(beamer)
Post on 03-Apr-2018
230 Views
Preview:
TRANSCRIPT
-
7/28/2019 Bessel- Legendre de 1(Beamer)
1/84
OutlinePower series
ExamplesExerciseAnswers
References
Special Functions: Bessel functions and Legendre
Polynomials
March 01 , 2013
1 / 3 8
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
2/84
OutlinePower series
ExamplesExerciseAnswers
References
Outline of Topics
1 Power seriesReal Analytic FunctionsSingular points
The Indicial equationComplete solutionMethod of reduction of order
2 Examples
3 Exercise
4 Answers
5 References
2 / 3 8
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
3/84
OutlinePower series
ExamplesExerciseAnswers
References
Real Analytic FunctionsSingular pointsComplete solutionMethod of reduction of order
Introduction
A power series is an infinite series of the form
m=0
am(x x0)m = a0 + a1(x x0) + a2(x x0)
2 + ...,
where the variable x, center x0, and the co-efficients a0, a1, ... arereal.
3 / 3 8
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
4/84
OutlinePower series
ExamplesExerciseAnswers
References
Real Analytic FunctionsSingular pointsComplete solutionMethod of reduction of order
The power series method is the standard basic method for solvinglinear differential equations with variable co-efficients. It givessolutions in the form of power series.
4 / 3 8
O li
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
5/84
OutlinePower series
ExamplesExerciseAnswers
References
Real Analytic FunctionsSingular pointsComplete solutionMethod of reduction of order
5 / 3 8
O tli
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
6/84
OutlinePower series
ExamplesExerciseAnswers
References
Real Analytic FunctionsSingular pointsComplete solutionMethod of reduction of order
A real function p(x) is called analytic at a point x = x0 if it can berepresented by a power series in powers of (x x0) with radius ofconvergence R > 0.
5 / 3 8
Outline
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
7/84
OutlinePower series
ExamplesExerciseAnswers
References
Real Analytic FunctionsSingular pointsComplete solutionMethod of reduction of order
Existence of power series solution
Theorem 2.1
If p(x), q(x) and r(x) are analytic at x = x0, then every solution of
y
+ p(x)y
+ q(x)y = r(x)
is analytic at x = x0 and can thus be represented by a power series
in powers of (x x0) with radius of convergence R > 0.
6 / 3 8
Outline
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
8/84
OutlinePower series
ExamplesExerciseAnswers
References
Real Analytic FunctionsSingular pointsComplete solutionMethod of reduction of order
Ordinary and singular points
y
+ p(x)y
+ q(x)y = 0 (1)A point x0
is said to be a regular(ordinary) point of the D.E. (1) if bothp(x) and q(x) are analytic at x = x0.
7 / 3 8
Outline
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
9/84
OutlinePower series
ExamplesExerciseAnswers
References
Real Analytic FunctionsSingular pointsComplete solutionMethod of reduction of order
Ordinary and singular points
y
+ p(x)y
+ q(x)y = 0 (1)A point x0
is said to be a regular(ordinary) point of the D.E. (1) if bothp(x) and q(x) are analytic at x = x0.
is said to be a singular point of the D.E. (1) if either p(x) orq(x) or both are not analytic at x = x0.
7 / 3 8
Outline
http://find/http://goback/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
10/84
OutlinePower series
ExamplesExerciseAnswers
References
Real Analytic FunctionsSingular pointsComplete solutionMethod of reduction of order
Classification of singular points:
regular(removable) singular points.
8 / 3 8
Outline
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
11/84
O tPower series
ExamplesExerciseAnswers
References
Real Analytic FunctionsSingular pointsComplete solutionMethod of reduction of order
Classification of singular points:
regular(removable) singular points.
irregular singular points.
8 / 3 8
Outline
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
12/84
Power seriesExamples
ExerciseAnswers
References
Real Analytic FunctionsSingular pointsComplete solutionMethod of reduction of order
Classification of singular points:
regular(removable) singular points.
irregular singular points.
A singular point x0 is called a regular singular point if both(x x0)p(x) and (x x0)
2q(x) are analytic.
8 / 3 8
Outline
http://find/http://goback/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
13/84
Power seriesExamples
ExerciseAnswers
References
Real Analytic FunctionsSingular pointsComplete solutionMethod of reduction of order
Classification of singular points:
regular(removable) singular points.
irregular singular points.
A singular point x0 is called a regular singular point if both(x x0)p(x) and (x x0)
2q(x) are analytic.On the other hand, if either (x x0)p(x) or (x x0)
2q(x) or both
are not analytic at x0, then x = x0 is called an irregular singularpoint.
8 / 3 8
Outline
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
14/84
Power seriesExamples
ExerciseAnswers
References
Real Analytic FunctionsSingular pointsComplete solutionMethod of reduction of order
Examples:
Any real number is a regular point of
y
+ (x2 + 1)y
+ (x3 + 2x2 + 3x)y = 0.
9 / 3 8
OutlineP i R l A l i F i
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
15/84
Power seriesExamples
ExerciseAnswers
References
Real Analytic FunctionsSingular pointsComplete solutionMethod of reduction of order
Examples:
Any real number is a regular point of
y
+ (x2 + 1)y
+ (x3 + 2x2 + 3x)y = 0.
1 are the only singular points of
(1 x2)y
+ 2y
+ (x3 3)y = 0.
9 / 3 8
OutlineP i R l A l ti F ti
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
16/84
Power seriesExamples
ExerciseAnswers
References
Real Analytic FunctionsSingular pointsComplete solutionMethod of reduction of order
Examples:
Any real number is a regular point of
y
+ (x2 + 1)y
+ (x3 + 2x2 + 3x)y = 0.
1 are the only singular points of
(1 x2)y
+ 2y
+ (x3 3)y = 0.
9 / 3 8
OutlinePower series Real Analytic Functions
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
17/84
Power seriesExamples
ExerciseAnswers
References
Real Analytic FunctionsSingular pointsComplete solutionMethod of reduction of order
Examples:
Any real number is a regular point of
y
+ (x2 + 1)y
+ (x3 + 2x2 + 3x)y = 0.
1 are the only singular points of
(1 x2)y
+ 2y
+ (x3 3)y = 0.
Both 1 are regular singular points.
9 / 3 8
OutlinePower series Real Analytic Functions
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
18/84
Power seriesExamples
ExerciseAnswers
References
Real Analytic FunctionsSingular pointsComplete solutionMethod of reduction of order
Examples:
The only singular points of
x3(x 1)y
+ 2(x 1)y
+ 5xy = 0
are
10/38
OutlinePower series Real Analytic Functions
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
19/84
Power seriesExamples
ExerciseAnswers
References
Real Analytic FunctionsSingular pointsComplete solutionMethod of reduction of order
Examples:
The only singular points of
x3(x 1)y
+ 2(x 1)y
+ 5xy = 0
are 0 and 1.
1 is an
10/38
OutlinePower series Real Analytic Functions
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
20/84
Power seriesExamples
ExerciseAnswers
References
Real Analytic FunctionsSingular pointsComplete solutionMethod of reduction of order
Examples:
The only singular points of
x3(x 1)y
+ 2(x 1)y
+ 5xy = 0
are 0 and 1.
1 is an regular singular point.
0 is an
10/38
OutlinePower series Real Analytic Functions
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
21/84
ExamplesExerciseAnswers
References
ySingular pointsComplete solutionMethod of reduction of order
Examples:
The only singular points of
x3(x 1)y
+ 2(x 1)y
+ 5xy = 0
are 0 and 1.
1 is an regular singular point.
0 is an irregular singular point.
10/38
OutlinePower series Real Analytic Functions
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
22/84
ExamplesExerciseAnswers
References
ySingular pointsComplete solutionMethod of reduction of order
Exercise: Classify the singular points of
1
x(x 1)y
+ y
+ (x2 1)y = 0.
11/38
OutlinePower series Real Analytic Functions
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
23/84
ExamplesExerciseAnswers
References
Singular pointsComplete solutionMethod of reduction of order
Exercise: Classify the singular points of
1
x(x 1)y
+ y
+ (x2 1)y = 0.
2 Legendres equation with n=1:
(1 x2)y
2xy
+ 2y = 0.
11/38
OutlinePower series Real Analytic Functions
S
http://find/http://goback/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
24/84
ExamplesExerciseAnswers
References
Singular pointsComplete solutionMethod of reduction of order
Exercise: Classify the singular points of
1
x(x 1)y
+ y
+ (x2 1)y = 0.
2 Legendres equation with n=1:
(1 x2)y
2xy
+ 2y = 0.
3
(x2 4)y
(2x 8)y = 0.
11/38
OutlinePower series
E lReal Analytic FunctionsSi l i
http://find/http://goback/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
25/84
ExamplesExerciseAnswers
References
Singular pointsComplete solutionMethod of reduction of order
The Form of the solution:
If x0 is a regular singular point of (1), then (1) admits at least one
nontrivial series solution of the form
y1(x) = (x x0)r
m=0
am(x x0)m
,
where r is a constant.
12/38
OutlinePower series
E lReal Analytic FunctionsSi l i t
http://find/http://goback/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
26/84
ExamplesExerciseAnswers
References
Singular pointsComplete solutionMethod of reduction of order
In fact r is a root of the quadratic equation obtained by equatingto zero the co-efficient of least power of x in the D.E. (1).
13/38
OutlinePower series
ExamplesReal Analytic FunctionsSingular points
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
27/84
ExamplesExerciseAnswers
References
Singular pointsComplete solutionMethod of reduction of order
In fact r is a root of the quadratic equation obtained by equatingto zero the co-efficient of least power of x in the D.E. (1).
The quadratic equation is called the indicial equation associatedwith the differential equation(1).
13/38
OutlinePower series
ExamplesReal Analytic FunctionsSingular points
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
28/84
ExamplesExerciseAnswers
References
Singular pointsComplete solutionMethod of reduction of order
In fact r is a root of the quadratic equation obtained by equatingto zero the co-efficient of least power of x in the D.E. (1).
The quadratic equation is called the indicial equation associatedwith the differential equation(1).
Remark: If x0 is an ordinary point , then r=0
13/38
OutlinePower series
ExamplesReal Analytic FunctionsSingular points
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
29/84
ExamplesExerciseAnswers
References
Singular pointsComplete solutionMethod of reduction of order
A second linearly independent solution y2(x) may be obtainedusing the method of reduction of order.
14/38
OutlinePower series
ExamplesReal Analytic FunctionsSingular points
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
30/84
ExamplesExerciseAnswers
References
Singular pointsComplete solutionMethod of reduction of order
Working rule:
Given a solution y1(x), let y2(x) be of the form
y2(x) = y1(x)v(x).
Substituting in
y
+ p(x)y
+ q(x)y = 0,
we get
15/38
OutlinePower series
ExamplesReal Analytic FunctionsSingular points
http://find/http://goback/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
31/84
pExerciseAnswers
References
g pComplete solutionMethod of reduction of order
Working rule:
Given a solution y1(x), let y2(x) be of the form
y2(x) = y1(x)v(x).
Substituting in
y
+ p(x)y
+ q(x)y = 0,
we get
u
+ u
2
y
1
y1+ p(x)
= 0, where u = v
,
-a first order differential equation(reduced order).
15/38
OutlinePower series
ExamplesReal Analytic FunctionsSingular points
http://find/http://goback/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
32/84
pExerciseAnswers
References
g pComplete solutionMethod of reduction of order
Working rule:
Given a solution y1(x), let y2(x) be of the form
y2(x) = y1(x)v(x).
Substituting in
y
+ p(x)y
+ q(x)y = 0,
we get
u
+ u
2
y
1
y1+ p(x)
= 0, where u = v
,
-a first order differential equation(reduced order).Separating the variables and integrating, we get
15/38
OutlinePower series
ExamplesReal Analytic FunctionsSingular points
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
33/84
ExerciseAnswers
References
Complete solutionMethod of reduction of order
Working rule:
Given a solution y1(x), let y2(x) be of the form
y2(x) = y1(x)v(x).
Substituting in
y
+ p(x)y
+ q(x)y = 0,
we get
u
+ u
2
y
1
y1+ p(x)
= 0, where u = v
,
-a first order differential equation(reduced order).Separating the variables and integrating, we get
v(x) =
1y21
e
p(x)dxdx.
15/38
OutlinePower series
ExamplesE i
Real Analytic FunctionsSingular pointsC l l i
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
34/84
ExerciseAnswers
References
Complete solutionMethod of reduction of order
Working rule:
Given a solution y1(x), let y2(x) be of the form
y2(x) = y1(x)v(x).
Substituting in
y
+ p(x)y
+ q(x)y = 0,
we get
u
+ u
2
y
1
y1+ p(x)
= 0, where u = v
,
-a first order differential equation(reduced order).Separating the variables and integrating, we get
v(x) =
1y21
e
p(x)dxdx.
Thus
y2(x) = y1(x)
1
y21
e
p(x)dx
dx.
15/38
OutlinePower seriesExamples
E i
Real Analytic FunctionsSingular pointsC l t l ti
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
35/84
ExerciseAnswers
References
Complete solutionMethod of reduction of order
Example1
y1(x) = cosx is a solution of
y
+ y = 0.
y2(x) = cosx v(x) v(x) =
16/38
OutlinePower seriesExamples
Exercise
Real Analytic FunctionsSingular pointsComplete solution
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
36/84
ExerciseAnswers
References
Complete solutionMethod of reduction of order
Example1
y1(x) = cosx is a solution of
y
+ y = 0.
y2(x) = cosx v(x) v(x) =
1cos2(x)
e
0dxdx = tanx
y2(x) = sinx.
16/38
OutlinePower seriesExamples
Exercise
Real Analytic FunctionsSingular pointsComplete solution
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
37/84
ExerciseAnswers
References
Complete solutionMethod of reduction of order
Example1
y1(x) = cosx is a solution of
y
+ y = 0.
y2(x) = cosx v(x) v(x) =
1cos2(x)
e
0dxdx = tanx
y2(x) = sinx.
Thus the complete solution is
c1cosx + c2sinx, where c1 and c2
are
16/38
OutlinePower seriesExamples
Exercise
Real Analytic FunctionsSingular pointsComplete solution
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
38/84
ExerciseAnswers
References
Complete solutionMethod of reduction of order
Example1
y1(x) = cosx is a solution of
y
+ y = 0.
y2(x) = cosx v(x) v(x) =
1cos2(x)
e
0dxdx = tanx
y2(x) = sinx.
Thus the complete solution is
c1cosx + c2sinx, where c1 and c2
are arbitrary constants.
16/38
OutlinePower seriesExamples
Exercise
Real Analytic FunctionsSingular pointsComplete solution
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
39/84
ExerciseAnswers
References
Complete solutionMethod of reduction of order
Example 2
y1(t) = 1 + t is a solution of
y
+2(t + 1)
t(t + 2)y
2
t(t + 2)y = 0.
Now y2(t) = (1 + t)v(t)
v(t) =
17/38
OutlinePower seriesExamples
Exercise
Real Analytic FunctionsSingular pointsComplete solution
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
40/84
ExerciseAnswers
References
Complete solutionMethod of reduction of order
Example 2
y1(t) = 1 + t is a solution of
y
+2(t + 1)
t(t + 2)y
2
t(t + 2)y = 0.
Now y2(t) = (1 + t)v(t)
v(t) =
1
(1 + t)2e
(
dt
t+
dt
t + 2)dt
=
1
(1 + t)21
t(t + 2)dt
=1
2ln
t
t + 2
+
1
t + 1.
17/38
OutlinePower seriesExamples
Exercise
Real Analytic FunctionsSingular pointsComplete solution
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
41/84
AnswersReferences
pMethod of reduction of order
Complete solution:
The general solution of the differential equation
y
+ p(x)y
+ q(x)y = 0
isy(x) = c1y1(x) + c2y2(x), where c1 and c2
are arbitrary constants.
18/38
OutlinePower seriesExamples
Exercise
http://find/http://goback/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
42/84
AnswersReferences
Example1:
Obtain the general solution of
y
+ y = 0
in powers of (x-0).
Solution:
Note that x=0 is a
19/38
OutlinePower seriesExamples
Exercise
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
43/84
AnswersReferences
Example1:
Obtain the general solution of
y
+ y = 0
in powers of (x-0).
Solution:
Note that x=0 is a regular point .
19/38
OutlinePower seriesExamples
ExerciseA
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
44/84
AnswersReferences
Example1:
Obtain the general solution of
y
+ y = 0
in powers of (x-0).
Solution:
Note that x=0 is a regular point . the differential equation admits a series solution of the form
19/38
OutlinePower seriesExamples
ExerciseA
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
45/84
AnswersReferences
Example1:
Obtain the general solution of
y
+ y = 0
in powers of (x-0).
Solution:
Note that x=0 is a regular point . the differential equation admits a series solution of the form say
y(x) =
m=0
am(x 0)m+r
.
19/38
OutlinePower seriesExamples
ExerciseAnswers
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
46/84
AnswersReferences
Substituting and simplifying,
20/38
OutlinePower seriesExamples
ExerciseAnswers
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
47/84
AnswersReferences
Substituting and simplifying,we get r=0 as a root of the indicial equation .
20/38
OutlinePower seriesExamples
ExerciseAnswers
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
48/84
AnswersReferences
Anda2m+1 =
a2m1
(2m + 1)(2m)
,
a2m+2 = a2m
(2m + 2)(2m + 1)
21/38
OutlinePower seriesExamples
ExerciseAnswers
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
49/84
AnswersReferences
Therefore
y(x) = a0
1
x2
2! +
x4
4! ...
+ a1
x
x3
3!
x5
5!
22/38
OutlinePower series
ExamplesExerciseAnswers
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
50/84
References
Therefore
y(x) = a0
1
x2
2! +
x4
4! ...
+ a1
x
x3
3!
x5
5!
i.e., y(x) = a0cosx + a1sinx, where a0 and a1 are arbitrary.
22/38
OutlinePower series
ExamplesExerciseAnswers
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
51/84
References
Example2:
Obtain the general solution of
x2y
+ (x2 3x)y
+ 3y = 0
in powers of (x-0).
Solution:
Note that x=0 is a
23/38
OutlinePower series
ExamplesExerciseAnswers
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
52/84
References
Example2:
Obtain the general solution of
x2y
+ (x2 3x)y
+ 3y = 0
in powers of (x-0).
Solution:
Note that x=0 is a regular singular point .
23/38
OutlinePower series
ExamplesExerciseAnswers
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
53/84
References
Example2:
Obtain the general solution of
x2y
+ (x2 3x)y
+ 3y = 0
in powers of (x-0).
Solution:
Note that x=0 is a regular singular point . the differential equation admits a series solution of the form
23/38
OutlinePower series
ExamplesExerciseAnswers
R f
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
54/84
References
Example2:
Obtain the general solution of
x2y
+ (x2 3x)y
+ 3y = 0
in powers of (x-0).
Solution:
Note that x=0 is a regular singular point . the differential equation admits a series solution of the form say
y(x) =
m=1
am(x 0)m+r
.
23/38
OutlinePower series
ExamplesExerciseAnswers
R f
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
55/84
References
Substituting and simplifying, we get
24/38
OutlinePower series
ExamplesExerciseAnswers
References
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
56/84
References
Substituting and simplifying, we get
(r(r 1) 3r + 3)a0xr
+
m=1
{[(m + r)(m + r 1) 3(m + r) + 3]am + [m + r 1]} am1xm+
The indicial equation is
24/38
OutlinePower series
ExamplesExerciseAnswers
References
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
57/84
References
Substituting and simplifying, we get
(r(r 1) 3r + 3)a0xr
+
m=1
{[(m + r)(m + r 1) 3(m + r) + 3]am + [m + r 1]} am1xm+
The indicial equation is
r2 4r + 3 = 0, ( as a0 = 0)
24/38
OutlinePower series
ExamplesExerciseAnswers
References
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
58/84
References
Substituting and simplifying, we get
(r(r 1) 3r + 3)a0xr
+
m=1
{[(m + r)(m + r 1) 3(m + r) + 3]am + [m + r 1]} am1xm+
The indicial equation is
r2 4r + 3 = 0, ( as a0 = 0)
The roots are
24/38
Outline
Power seriesExamples
ExerciseAnswers
References
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
59/84
References
Substituting and simplifying, we get
(r(r 1) 3r + 3)a0xr
+
m=1
{[(m + r)(m + r 1) 3(m + r) + 3]am + [m + r 1]} am1xm+
The indicial equation is
r2 4r + 3 = 0, ( as a0 = 0)
The roots are r=3 and 1.
24/38
Outline
Power seriesExamples
ExerciseAnswers
References
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
60/84
Substituting r=3 and equating the co-efficients of xm+r on eithersides, we get the following recurrence relation:
25/38
Outline
Power seriesExamples
ExerciseAnswers
References
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
61/84
Substituting r=3 and equating the co-efficients of xm+r on eithersides, we get the following recurrence relation:
am =(m + 2)am1
(m + 3)(m 1) + 3
, m 1.
25/38
Outline
Power seriesExamples
ExerciseAnswers
References
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
62/84
Substituting r=3 and equating the co-efficients of xm+r on eithersides, we get the following recurrence relation:
am =(m + 2)am1
(m + 3)(m 1) + 3
, m 1.
Thus
a1 = a0 , a2 =
25/38
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
63/84
Outline
Power seriesExamples
ExerciseAnswers
References
-
7/28/2019 Bessel- Legendre de 1(Beamer)
64/84
Substituting r=3 and equating the co-efficients of xm+r on eithersides, we get the following recurrence relation:
am =(m + 2)am1
(m
+ 3)(m 1) + 3
, m 1.
Thus
a1 = a0 , a2 =1
2a0 , a3 =
a03
, a4 =a0
4, ....
y1(x) = a0x3
ex
.
25/38
Outline
Power seriesExamples
ExerciseAnswers
References
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
65/84
To find y2(x), apply the method of reduction of order with a0 = 1.
Lety2 = y1v.
Then
v(x) =
ex
x3dx.
26/38
Outline
Power seriesExamples
ExerciseAnswers
References
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
66/84
To find y2(x), apply the method of reduction of order with a0 = 1.
Lety2 = y1v.
Then
v(x) =
ex
x3dx.
i.e., v =1
2x2
1
x+
1
2lnx +
1
6x +
1
48x2 + ....
26/38
Outline
Power seriesExamples
ExerciseAnswers
References
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
67/84
To find y2(x), apply the method of reduction of order with a0 = 1.
Lety2 = y1v.
Then
v(x) =
ex
x3dx.
i.e., v =1
2x2
1
x+
1
2lnx +
1
6x +
1
48x2 + ....
y2(x) = (1
2x
1
2x2 +
3
4x3 ...) +
1
2x3exlnx.
26/38
Outline
Power seriesExamples
ExerciseAnswers
References
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
68/84
To find y2(x), apply the method of reduction of order with a0 = 1.
Lety2 = y1v.
Then
v(x) =
ex
x3dx.
i.e., v =1
2x2
1
x+
1
2lnx +
1
6x +
1
48x2 + ....
y2(x) = (1
2x
1
2x2 +
3
4x3 ...) +
1
2x3exlnx.
Thusy(x) = c1y1(x) + c2y2(x), where c1 and c2
are arbitrary constants.
26/38
Outline
Power seriesExamples
ExerciseAnswers
References
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
69/84
Have some Patience?
27/38
Outline
Power seriesExamples
ExerciseAnswers
References
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
70/84
Have some Patience?
yes...1 Obtain the general solution of
x(x + 1)y
(2x + 1)y = 0
in powers of (x-0).2 Obtain the general solution of
(1 x2)y
2xy
+ 2y = 0
in powers of (x-0).3 Obtain a series solution of
(8x2)y
+ 10xy
(1 + x)y = 0
about 0.27/38
Outline
Power seriesExamples
ExerciseAnswers
References
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
71/84
Still Have Patience?
28/38
Outline
Power seriesExamples
ExerciseAnswers
References
S ll H P ?
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
72/84
Still Have Patience?
Good...4 Obtain a series solution of
(4x2)y
+ 2x(2 x)y
(1 + 3x)y = 0
about 0.5 Obtain the general solution of
xy
+ 3y
+ 4x3y = 0
in powers of (x-0).
6 Obtain the general solution of
(x 1)2y
+ (x 1)y
4y = 0
about x=1. (Hint: put x-1=t)
28/38
Outline
Power seriesExamples
ExerciseAnswers
References
S ill ??
http://goforward/http://find/http://goback/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
73/84
Still ??
29/38
Outline
Power seriesExamples
ExerciseAnswers
References
S ill ??
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
74/84
Still ??
Very Good...
7 Obtain the general solution of
(1 + x)x2y
(1 + 2x)xy
+ (1 + 2x)y = 0
about x=0.
29/38
Outline
Power seriesExamples
ExerciseAnswers
References
A
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
75/84
Answers:
Try yourself.
30/38
Outline
Power seriesExamples
ExerciseAnswers
References
A
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
76/84
Answers:
Try yourself.
1
y(x) = a1x + a2x2
.
2
y(x) = a1x + a0
1 x2
1
3x4
1
5x6 + ...
.
30/38
Outline
Power seriesExamplesExerciseAnswers
References
Ans ers
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
77/84
Answers:
3 r=1/4,-1/2;
y1(x) =
m=0
amxm+
1
4 ,
where am =a0
2mm!7.11.15....(4m 1)(4m + 3),
m = 1, 2, ... a0 is arbitrary.
31/38
Outline
Power seriesExamplesExerciseAnswers
References
Answers:
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
78/84
Answers:
4 r=1/2,-1/2;
y1(x) =
m=1
a0
2m1(m 1)!x
m1
2 , where a0 is a arbitrary constant.
32/38
Outline
Power seriesExamplesExerciseAnswers
References
Answers:
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
79/84
Answers:
5 r=0,-2;y1(x) = a0x
2 sin(x2),
v(x) =cot(x2)
2; y2(x) =
x2cos(x2)
2.
33/38
Outline
Power seriesExamplesExerciseAnswers
References
Answers:
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
80/84
Answers:
6 r=2,-2;
y1(t) = a0t2
,
v(t) =t4
4; y2(x) =
t2
4
y(x) = c1(x 1)2 + c2(x 1)
2.
34/38
Outline
Power seriesExamplesExerciseAnswers
References
Answers:
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
81/84
Answers:
7 r=1,1; y1(x) = a0x,
v(x) = x + ln x; y2(x) = x(x + ln x).
35/38
Outline
Power seriesExamplesExerciseAnswers
References
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
82/84
For more details , you can click on Series Solutions of DifferentialEquations.
36/38 Outline
Power seriesExamplesExerciseAnswers
References
http://www.math.ualberta.ca/~xinweiyu/334.1.10f/DE_series_sol.pdfhttp://www.math.ualberta.ca/~xinweiyu/334.1.10f/DE_series_sol.pdfhttp://www.math.ualberta.ca/~xinweiyu/334.1.10f/DE_series_sol.pdfhttp://www.math.ualberta.ca/~xinweiyu/334.1.10f/DE_series_sol.pdfhttp://www.math.ualberta.ca/~xinweiyu/334.1.10f/DE_series_sol.pdfhttp://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
83/84
Erwin Kreyzig., Advanced Engineering Mathematics.
B.S. Grewal, Higher Engineering Mathematics.
37/38 Outline
Power seriesExamplesExerciseAnswers
References
http://find/ -
7/28/2019 Bessel- Legendre de 1(Beamer)
84/84
Thank You
38/38
http://find/
top related