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Power series solution of Differential equations about ordinary points

Institute of Lifelong Learning, University of Delhi

Subject: Differential Equations & Mathematical Modeling -III

Lesson: Power series solutions of Differential Equations

about ordinary points

Lesson Developer: Nisha Bohra and Ankit Gupta

College/Department: Sri Venkateswara College, University of

Delhi Ramjas College, University of Delhi


Institute of Lifelong Learning, University of Delhi Page | 2

Table of Contents

1. Learning Outcomes

2. Introduction

3. Basic Concepts and Results

3.1. Second Order homogenous linear differential

equation

3.2. Power Series and its Radius of Convergence

3.3. Power Series method for solving a differential

eqation

Exercises

4. Ordinary and Singular points

Exercises

5. Series Solution near Ordinary Point

Exercises

6. Legendre’s Equation

Summary

References



1. Learning Outcomes

After reading this lesson reader will be able to understand the following

Second order homogenous linear differential equation

Power Series

Radius of convergence

Ordinary points, singular points

Solution about an ordinary point

Legendre’s equation

2. Introduction

We have studied earlier that the problem of solving a homogenous linear

differential equation with constant coefficients can be simplified to the algebraic

problem of finding the roots of characteristic equation. However there is no

parallel system for solving linear differential equation with variable coefficients.

Thus we must seek other techniques for the solutions of these equations and this

chapter is devoted to the methods of obtaining solutions in infinite series form.

3. Basic Concepts and Results

3.1. Second order homogenous linear differential equation

A differential equation of the form

2

20,

d y dyA x B x C x y

dx dx

where the coefficients A, B and C are continuous real functions of x on an interval

I and A(x) is not identically zero on I, is called a second order homogenous linear

differential equation.

3.2. Power Series and its radius of convergence

Definition. A power series about a is an infinite series of the form

2

0 1 20

...n

nn

C x a C C x a C x a (3.1)

where nC ’s are constants.

For the sake of simplicity of our notation, we shall treat only the case when a=0.

This is no loss of generality, since the translation 'x x a reduces a power

series around the point to a power series around zero. Thus, we shall mainly

study a series of the form

20 1 2

0

...nn

n

C x C C x C x (3.2)



Even though the series (3.2) is defined over all of ℝ, it is not to be expected that

the series will converge for all x in ℝ. For example, the geometric series

2

0

1 ...n

n

x x x converges for 1x and the series

0

/ !n

n

x n converges

for all x in ℝ as can be seen by Ratio test.

Radius of Convergence (R) of a power series

Definition. For a given power series

0

n

nn

C x , we define the number R,

0 R , by

1/1

limsupn

nC

R, then

a) If x R , the series converges absolutely.

b) If x R , the series diverges.

c) If 0 r R , then the series converges uniformly on :x x r .

The number R is called the radius of convergence and the open interval

,R R is known as interval of convergence of the power series nnC x .

Remark. The radius of convergence R of the series

0

Nn

nn

C x is also given by

1

lim n

nn

C

C, provided this limit exists. If R , we say series converges for all x

and if 0R , the series diverges for all 0x .

Example 1. If

0

1

3

n

nn

ny x x

then

1

lim n

nn

CR

C

Converges absolutely Diverges Diverges



11 3lim

23

n

nn

n

n

13lim

2n

n

n

3

Hence the series converges for | | 3x and diverges for | | 3x .

We now state some basic results on Power series without proof.

Theorem 1. Term wise differentiation of Power series

Let a function f has the following power series representation,

0 10

... ...n n

n nn

f x C x C C x C x

If it converges on the open interval I, then we say, f is differentiable on I and

1 21 2 3

1

2 3 ...nn

n

f x nC x C C x C x

at each point of I. Both series have the same radius of convergence.

Theorem 2. Identity principle

If

0 0

n nn n

n n

a x b x for every x in an open interval I, then an = bn for all 0.n

In particular, if

0

0nn

n

a x for all x in I, it follows that 0 0na n .

3.3. Power series method for solving a differential equation

In this section, we illustrate the power series method for finding the

solution of a differential equation.

Given a differential equation, we will assume that it has a solution of the form

0

( ) n

nn

y x C x .

Then, by substituting the above power series in the given differential equation,

we will find the coefficients 0 1 2, , ,C C C …. or the set of conditions which must be

satisfied by them.

Let us try to understand this method with the help of an example.

Example 2. Solve the equation

3 2 0dy

x ydx

. (3.3)



Solution. Let

0

( ) n

nn

y x C x be the solution of (3.3).

Differentiating term by term we obtain

1

1

n

nn

dynC x

dx.

Substituting the values of y and y’ into the differential equation (3.3), we have

1

1 0

3 2 0n n

n nn n

x nC x C x

Since x is independent of the index of summation n, we may rewrite this as

1

1 1 0

3 2 0n n n

n n nn n n

nC x nC x C x .

In order to make the exponent of x same in all the three summations, we shall

rewrite second summation.

Consider the second summation

1

1

3 n

nn

nC x .

To make the exponent n, we first replace the present exponent (n-1) by a new

variable m. That is, we let m = n-1. Then n = m+1, and since m=0 for n=1, the

summation takes the form

10

3 ( 1) m

mm

m C x . Now since the variable of

summation is merely a “dummy” variable, we may replace m by n to write the

second summation as

10

3 ( 1) n

nn

n C x . (3.4)

Replacing second summation by its equivalent form (3.4), we get

10 0 0

3 1 2 0n n nn n n

n n n

nC x n C x C x

.

10

3 1 2 0n

n n nn

nC n C C x .

The identity principle gives

13 1 2 0

n n nnC n C C , n≥ 0.

This yields the recurrence relation

1

2

3 1n n

nC C

n

for 0n .

Put n = 0, 1, 2, we get



1 0

2

3C C , 2 1 02

3 3

3.2 3C C C ,

3 03

4

3C C

Generalizing the above pattern, we can write

0

1

3n n

nC C

, 1n

Hence power series solution of (3.3) is

00

1

3

n

nn

ny x C x

. (3.5)

The radius of convergence of above power series is 3 3

lim 32n

nR

n

. Hence

(3.5) converges absolutely if and diverges for .

Exercise

Find the radius of convergence of following power series.

1.

0

1 n

n

xn

2.

0 !

nn

n

nx

n

3.

2

0

( 1)

(2 )!

nn

n

xn

4.

2

0

( 1)nn

n

x

Find a power series solution of following differential equations. Also, determine

the radius of convergence of the resulting solution series.

1. 2 3 0dy

ydx

2. 2 0dy

xydx

3. 4dy

ydx

4. 2dyx y

dx

4. Ordinary and Singular Points

In Example 2 of previous section, we assumed that the differential

equation has a power series solution. However it is not always true that the

differential equation possesses a solution of this form. Hence the natural question

is “under what conditions this assumption is actually valid?” In order to answer



this important question concerning the existence of a power series solution, we

shall first introduce a few basic definitions.

Definition. A real valued function f defined on an interval I containing the point a

is said to be analytic at x = a if its Taylor series about a,

0 !

nn

n

f ax a

n exists

and converges to f x for all x in some open interval containing a.

Value Addition: Note

All polynomial functions, exponential function and trigonometric functions like

sinx and cosx are analytic everywhere. A rational function is analytic except at

points where denominator is zero. For example, the rational function defined by

21 / 3 2x x is analytic everywhere except at x = 1 and x = 2.

Consider the second order homogeneous linear differential equation of the form

2

20

d y dyA x B x C x y

dxdx (4.1)

where the coefficients A, B and C are analytic functions of x.

Rewriting equation (4.1) in the form

2

20

d y dyP x Q x y

dxdx (4.2)

with leading coefficient 1 and with /P x B x A x and /Q x C x A x .

Equation (4.2) is called equivalent normalized form of equation (4.1).

Note that P x and Q x will fail to be analytic at points where A x is zero.

Definition. A point x = a is called an ordinary point of the differential equation

(4.1) and the equivalent equation (4.2) if both the functions P(x) and Q(x) are

analytic at x = a. If either (or both) of these functions are not analytic at

x = a, then a is called singular point.

Example 1. Consider the differential equation

2

2

22 0

d y dyx x y

dxdx.

Here P x x and 2 2Q x x .

Both functions P and Q are polynomial functions and so they are analytic

everywhere. Thus every point is an ordinary point of the given differential

equation.

Example 2. Consider the differential equation

2

2

11 0

d y dyx x y

dx xdx.



We first express the above differential equation in normalized form, thereby

obtaining

2

2

10

1 1

d y x dyy

dxx x xdx.

Here,

1

xP x

x and

1

1Q x

x x.

The function P is not analytic at x = 1 and Q is not analytic at the points

x = 0 and 1. Thus x = 0 and x = 1 are singular points of given differential

equation. All other points are ordinary points.

Example 3: Consider the differential equation

2

2

2sin 0

d y dyx x x y

dxdx.

Then the point x = 0 is an ordinary point because the function

3 5sin 1( ) ...

3! 5!

x x xP x x

x x

2 4

1 ...3! 5!

x x

is clearly analytic at x = 0 and Q x x being a polynomial function is also

analytic at x = 0.

Exercise

In problems 1 to 4, determine the ordinary and singular points.

1. 2

23 2 0

d y dyx x y

dxdx.

2. 2

2

21 6 12 0

d y dyx x y

dxdx.

3. 2

2

21 0

d y dyx x xy

dxdx.

4. 2

2

21 4 12 0

d y dyx x y

dxdx.

5. Series solution near ordinary points

We now state a theorem concerning the existence of power series solutions of the

form

0

n

nn

C x a .



Theorem 3. Suppose that a is an ordinary point of the differential equation

2

20

d y dyA x B x C x y

dxdx (5.1)

i.e., the functions /P B A and /Q C A are analytic at x = a. Then equation

(5.1) has two linearly independent solutions, each of the form

0

n

nn

y x C x a (5.2)

and these power series solutions converges in some interval x a R about a (R

> 0). We shall omit the proof of this important theorem.

Remark. The radius of convergence R of such series solution is at least as large

as the distance of a from the nearest singular point (real or complex) of the

equation (5.1).

Example 1. Determine the guaranteed radius of convergence of a series solution

of

2

2 2

2( 9) 0

d y dyx x x y

dxdx

in powers of x. Repeat for a series in powers of (x-4).

Solution. We first write the differential equation in normalized form.

2 2

2 2 20

( 9) ( 9)

d y x dy xy

dxdx x x.

Here

2 9

xP x

x and

2

2 9

xQ x

x. Thus, the only singular points of

given differential equation are . The distance of both from 0 is 3, so the

radius of convergence of series solution

0

n

nn

C x is at least 3. Since, the distance

of both singular point from 4 is 5, so a series solution of the form



0

( 4)n

nn

C x has radius of convergence at least 5.

Remark. Theorem 3 gives us only sufficient condition for the existence of power

series solutions of the differential equation (5.1). In Example 2 of section 4, we

observed that x = 0 and x = 1 are the only singular points of the given

differential equation. Thus the differential equation has two linearly independent

solutions of the form (5.2) about any point x ≠ 0 or 1. However we are not

assured that any solution of the form

0

nn

n

C x about the singular point x = 0 or

any solution of the form

0

1n

nn

C x about the singular point 1.

Let us look at some examples based on finding the power series solution

of given differential equation about an ordinary point.

Example 2. Find the general solution of the differential equation

2

2

22 0

d y dyx x y

dxdx (5.3)

in powers of x (that is about a = 0).

Solution. Clearly a = 0 is an ordinary point of (5.3).

We assume that

0

( ) n

nn

y x C x is a solution of (5.3).

Differentiating term by term, we obtain

1

1

nn

n

dynC x

dx and

2

2

22

1 nn

n

d yn n C x

dx.

Substituting these values in equation (5.3), we obtain

2 2

2 1 0

1 2 0n n nn n n

n n n

n n C x nC x x C x .

2 2

2 1 0 0

1 2 0n n n nn n n n

n n n n

n n C x nC x C x C x .

To make the exponent of x same in all the summations, we shift the index by 2

in the first sum, replacing n by (n + 2) and using the initial value n = 0. This

gives

2

20 1 0 0

1 2 2 0n n n n

n n n nn n n n

n n C x nC x C x C x

Also in the third sum, we replace n by (n – 2) to get



2 20 1 2 0

1 2 2 0n n n n

n n n nn n n n

n n C x nC x C x C x (5.4)

Since range of various summations is not the same. The common range is from 2

to . We now write the terms in each summation for n=0 and n=1 separately and

we continue to use the "sigma" notation for the remainder of each such

summation. Thus equation (5.4) reduces to

2 3 2 1 2 02 2 2

2 6 2 1 2n n n

n n nn n n

C C x n n C x C x nC x C x C

12

2 2 0nn

n

C x C x .

This gives

0 2 1 3 2 22

2 2 3 6 2 1 2 0n n n

n

C C C C x n n C n C C

Equating coefficient of each power of x in the left to zero, we get

0 22 2 0C C

1 33 6 0C C

2 22 1 2 0, 2n n nn n C n C C n (5.5)

2 0C C

& 3 1

1

2C C

The condition (5.5) is called recurrence formula. It enables us to express each

coefficient 2nC for 2n in terms of the previous coefficients

nC and2nC , thus

giving

2

2

2

1 2

n n

n

n C CC

n n

, 2n

Putting n = 2, we get 4 0

1

4C C .

For n = 3,

3 15

5

20

C CC and then using the value of

3C , we have 5 1

3

40C C

In this way, we can express each even coefficient in terms of C0 and each odd

coefficient in terms of C1.

Substituting the values of 2 3 4, ,C C C & 5C in the assumed solution, we have

2 3 4 5

0 1 2 1 0 1

1 1 3( ) ...

2 4 40y x C C x C x C x C x C x



Collecting terms in 0

C and 1

C , we have

2 4 3 5

0 1

1 1 3( ) 1 ... ...

4 2 40y x C x x C x x x

(5.6)

which gives the solution of the differential equation (5.3) in powers of x.

The two series in the parenthesis are the power series expansions of two

linearly independents solutions of (5.3) and C0, C1 are arbitrary constants. Thus

(5.6) represents the general solution of (5.3) in powers of x.

Example 3. Find the general solution in powers of x of

2

2

24 3 0.

d y dyx x y

dxdx (5.7)

Also, find the particular solution using conditions y(0) = 4, ' 0 1y .

Solution. Clearly a = 0 is an ordinary point of (5.7). We assume that

0

( ) n

nn

y x C x (5.8)

is a solution of (5.7).

Differentiating (5.8) term by term, we get

1

1

n

nn

dynC x

dx

2

2

22

1 n

nn

d yn n C x

dx

Substituting these values in (5.7) yields

2 2 1

2 1 0

4 1 3 0n n n

n n nn n n

x n n C x x nC x C x .

2

2 2 1 0

1 4 1 3 0n n n n

n n n nn n n n

n n C x n n C x nC x C x .

We can change the initial value from n=2 and n=1 to n=0 in the first and third

summation without affecting the sum. Also, by replacing n with n + 2 and using

the initial value n = 0, we shift the index of summation in the second term by +2.

This gives

20 0 0 0

1 4 2 1 3 0n n n n

n n n nn n n n


20

1 4 2 1 3 0nn n n n

n

n n C n n C nC C x

.



20

( 1 3 1) 4 2 1 0n

n nn

n n n C n n C x .

The identity principle yields

2

21 4 2 1 0n nn C n n C , n≥0.

2

2

1

4 2 1

n

n

n CC

n n

1

4 2

nn C

n, n≥0 (5.9)

(5.9) gives the recurrence relation for 0n . With n = 0, 2 & 4, we get

02

4.2

CC , 2

4

3

4.4

CC = 0

2

3

4 .2.4

C

46

5

4.6

CC = 0

3

3.5

4 .2.4.6

C

Generalizing the above pattern, we have

2 0

1.3.5 ... 2 1

4 .2.4 ... 2n n

nC C

n

03

1.3.5 ... 2 1

2 . !n

nC

n

, n ≥ 1

With n = 1, 3 & 5 in (5.9), we get

3 1

2

4.3C C , 5 3 12

4 2.4

4.5 4 .3.5C C C

and 57 13

6 2.4.6

4.7 4 .3.5.7

CC C


2 1 1

2.4.6 ... 2

4 .1.3.5 ... 2 1n n

nC C

n

1

!, 1

2 .1.3.5...(2 1)n

nC n

n

Using these values in the assumed solution (5.8), we get



2 2 1

0 131 1

1.3.5 ... 2 1 !1

2 . ! 2 .1.3.5 ... 2 1

n n

n nn n

n ny x C x C x x

n n

OR

2 4 3 5

0 1

1 3 1 11 ... ...

8 128 6 30y x C x x C x x x .

Since 00y C & 10y C .

Using initial conditions, we have C0 = 4 and C1 = 1.

Hence, the particular solution of (5.7) is

2 4 3 51 3 1 14 1 ... ...

8 128 6 30y x x x x x x

2 3 4 51 1 3 14 ...

2 6 32 30x x x x x

We note that singular points of equation (5.7) are 2 , so the radius of

convergence of above series is at least 2.

Translated Series Solutions.

Suppose we are required to solve a differential equation with initial values

specified at a point a, i.e. y a and y a are given. To find a particular solution

with given initial values, we will need a general solution of the form

0

n

nn

y x C x a

.

Because then 0y a C and 1y a C .

And thus we have the values of C0 & C1 in terms of the y a and 'y a .

Therefore, to solve an initial value problem, we require a general solution

centered at the point at which initial conditions are given. Let us look at some

examples based on this.

Example 4. Find a power series solution of the initial value problem

2

2

21 3 0

d y dyx x xy

dxdx (5.10)

0 4y , 0 6y

Solution. We observe that all points except 1x are ordinary points for the

differential equation (5.10).

Since the initial values of y and its first derivative are prescribed at x = 0, we

assume



0

( ) n

nn

y x C x as the general solution of (5.10).

Differentiating term by term, we obtain

1

1

nn

n

dynC x

dx

,

2

2

22

1 n

nn

d yn n C x

dx

Substituting the values of ,y ,dy

dx and

2

2

d y

dx in (5.10), we have

2 1

2 2 1 0

1 1 3 0n n n n

n n n nn n n n


To make the exponent of x same in each term, we replace n by 2n in second

sum and n by 1n in fourth sum we get,

2 12 0 1 1

1 2 1 3 0n n n n

n n n nn n n n


The common range of these terms is form 2 to . We write the terms for n=0

and n=1 in each sum separately and thus above equation takes the form

2 3 2 12 2 2

1 2 6 2 1 3 3n nn n n

n n n

n n C x C C x n n C C x nC x

0 12

0n

nn

C x C x .

Combining like powers of x, this takes the form

2 0 1 3

2 ( 3 6 )C C C C x

2 12

2 1 1 3 0n

n n n nn

n n C n n C nC C x

using identity principle, we get

C2 = 0, 0 1 33 6 0C C C and

2 12 1 2 0n n nn n C n n C C

3 0 1

1 1

6 2C C C

The recurrence formula gives

1

2

2

2 1

n n

n

n n C CC

n n , 2n .



Put n = 2, 2 14 1

8 1

12 12

C CC C

Put n = 3, 3 25 0 1

15 1 3

20 8 8

C CC C C

.

Substituting the values of 2 3 4, 5, ,C C C C into the assumed solution of (5.10), we

get

3 4 50 01 1 10 1

3( ) ...

6 2 12 8 8

C CC C Cy x C C x x x x

OR

43 5 3 5

0 1

1 1 1 3( ) 1 ... ...

6 8 2 12 8

xy x C x x C x x x

This represents the general solutions of the differential equation (5.10).

0

0 4 4y C , 1

0 6 6y C .

Hence the particular solution is given by

3 5 3 41 1 1 1( ) 4 1 ... 6 ...

6 8 2 12y x x x x x x

3 4 511 1 11( ) 4 6 ...

3 2 4y x x x x x

Example 5. Solve the initial value problem

2

2

22 3 3 1 0

d y dyt t t y

dtdt (5.11)

1 4y and 1 1y

Solution: Since the initial conditions are given at t = 1, we will assume a power

series solution of the form 0

1n

nn

y t C t

of (5.11).

Also, we note that t = 1 is an ordinary point of (5.11).

Here, we will not substitute the assumed series solution in (5.11) as done in other

problems, instead we first substitute x= (t-1) in (5.11), so that we end up finding

a series of the form 0

nn

n

C x

.

Now the substitution 1x t changes 2 2 3t t to 2 21 2 1 3 4x x x

and



.dy dy dx dy

dt dx dt dx

2 2

2 2.

d y d dy dx d y

dx dx dtdt dx

Hence, equation (5.11) changes to

2

2

24 3 0

d y dyx x y

dxdx with initial conditions 0 4y & 0 1y

corresponding to t = 1.

This is the initial value problem we solved in Example 3 of this section.

The particular solution obtained in Example 3 was

2 3 41 1 3( ) 4 ...

2 6 32y x x x x x

Replacing x by 1t , we have the required particular solution

2 3 41 1 3

4 1 1 1 1 ...2 6 32

y t t t t t (5.12)

Also, note that the solution in terms of x converges for | | 2x

Hence (5.12) converges for | 1| 2t or for 1 3t .

Exercise

Find power series solution in powers of x of differential equations from 1–6. Also,

find the radius of convergence in each case.

1. 2

20

d y dyx y

dxdx

2. 2

2

22 4 2 0

d y dyx x y

dxdx

3. 2

2

22 1 0

d y dyx x y

dxdx

4. 2

2

21 6 4 0

d y dyx x y

dxdx

5. 2

23 2 0

d y dyx x y

dxdx

6. 2

2

21 6 12 0

d y dyx x y

dxdx

Find power series solution of the initial value problems in 7–10.



7. 2

20

d y dyx y

dxdx; 0 1y , 0 0y

8. 2

2

21 2 2 0

d y dyx x y

dxdx; 0 1y , 0 1y

9. 2

2

21 2 0

d y dyx x xy

dxdx; 0 2y , 0 3y

10. 2

22 0

d y dyx y

dxdx; 0 1y , 0 0y

Solve the initial value problems in 11 – 14. Also, find the interval of convergence.

11. 2

21 0

d y dyx y

dxdx; 1 2y , 1 0y

12. 2

2

22 6 1 4 0

d y dyx x x y

dxdx; 1 0y , 1 1y

13. 2

2

26 10 4 3 6 0

d y dyx x x y

dxdx; 3 2y , 3 0y

14. 2

2

26 3 9 3 0

d y dyx x x y

dxdx; 3 0y , 3 2y

6. The Legendre’s Equation

The second order linear differential equation

2

2

21 2 1 0

d y dyx x y

dxdx (6.1)

is called the Legendre’s equation of order , where the real no. satisfies the

inequality > –1.

A solution to this equation is called a Legendre function. This differential equation

has wide range of applications.

Value addition: Note



Adrien Marie Legendre invented Legendre Polynomials (the contribution for

which he is best remembered) in the context of gravitational attraction of

ellipsoids. Legendre was a fine French Mathematician who suffered the

misfortune of seeing most of his best work –in elliptic integrals, number theory

and the method of least squares- superseded by the achievements of younger

and abler men. For instance, he devoted 40 years to the study of elliptic

integrals, and his two volume treatise on the subject had scarcely appeared in

print before the discoveries of Abel and Jacobi revolutionized the field.

Legendre was remarkable for the generous spirit with which he repeatedly

welcomed newer and better work that made his own obsolete.

Writing the Legendre equation in normalized form, we have

2

2 2 2

120

1 1

d y x dyy

dxdx x x.

Clearly, x=0 is an ordinary point and x = 1,-1 are its only singular points. The

Legendre equation has two linearly independent solutions expressible as power

series in x with radius of convergence at least 1.

Let

0

( ) n

nn

y x C x be the power series solution of (6.1)

Differentiating term by term, we get

1

0

n

nn

dynC x

dx and

2

2

20

1 n

nn

d yn n C x

dx.

Substituting these values in (6.1), we have

2

2 2 1

1 1 2 1n n n

n n nn n n

n n C x n n C x nC x0

0nn

n

C x

To make the exponent of x same in all summations, we replace n by n + 2 in first

summation. Also we shift the index of sum from n=2 to n=0 in second term and

from n=1 to n=0 in third term. This gives

20 0 0

2 1 1 2 1n n n

n n nn n n

n n C x n n C x nC x

0

0.n

nn

C x

20

2 1 1 2 1 0n

n n n nn

n n C n n C nC C x

By identity principle

22 1 1 2 1 0

n nn n C n n n C .

Thus,

2

2

1

2 1n n

n nC C

n n



1

2 1n

n nC

n n

1

2 1n

n nC

n n , 0n

Put n= 0, 1, 2, 3, we have

2 0

1

2!C C

,

3 1

1 2

3!C C

4 0

2 1 3

4!C C

,

5 1

1 3 2 4

5!C C


2 0

1 2 ... 2 2 1 3 ... 2 1

2 !

n

n

n nC C

n

2 1 1

1 1 3 ... 2 1 2 4 ... 2

2 1 !

n

n

n nC C

n

Let

2

2 ... 2 2 1 3 ... 2 1

2 !n

n na

n

2 1

1 3 ... 2 1 2 4 ... 2

2 1 !n

n na

n

.

Then, 2 2 01n

n nC a C , n≥1 and 2 1 2 1 11n

n nC a C , n≥1.

We have two power series solutions of Legendre's equation of order which are

linearly independent

21 0 2

0

1n n

nn

y x C a x

and 2 12 1 2 1

0

1n n

nn

y x C a x

And the general solution is linear combination of 1( )y x and 2 ( )y x .

Now, suppose m , a non-negative integer.

If m is even, then 2 0na when 2n > m.

In this case, 1y x is a polynomial of degree m and 2y x is an infinite series

(non-terminating).

If m is an odd integer, then 2 1 0na when 2n + 1 > m.



In this case, 2y x is a polynomial of degree m and 2 ( )y x is a non-terminating

infinite series.

Thus in either case, one of the two solutions is a polynomial or other is a

nonterminating series. With an appropriate choice of the arbitrary constants 0C

(m even) and 1C (m odd), the Legendre’s equation has polynomial solution. The

mth degree polynomial solution of Legendre’s equation of order m,

2(1 ) '' 2 ' ( 1) 0,x y xy m m y

is denoted by ( )mP x and is called the Legendre polynomial of degree m. It is

customary to choose an arbitrary constant so that the coefficient of nx in ( )n

P x is

2(2 !) / [2 ( !) ].nn n It then turns out that

2

0

( 1) (2 2 )!( ) ,

2 !( )!( 2 )!

kNn k

n nk

n kP x x

k n k n k

where / 2N n , the integral part of / 2n . The first six Legendre polynomials

are

0( ) 1P x ,

1( ) ,P x x 2

2

1( ) (3 1),

2P x x 3

3

1( ) (5 3 ),

2P x x x

4 2

4

1( ) (35 30 3),

8P x x x 5 3

5

1( ) (63 70 15 )

8P x x x x .

Graphs ( )n

y P x of the Legendre polynomial for 0,1,2,3,4,5.n



Summary

In this lesson we have defined and emphasized on the following:

Second order homogenous linear differential equation

Power Series

Radius of convergence

Ordinary points, singular points

Solution about an ordinary point by power series method

Legendre's equation

References

[1] C.H. Edwards and D.E. Penny, Differential Equations and boundary Value

Problems: Computing and Modeling, Pearson Education, India, 2005.

[2] S. L. Ross, Differential equations, 3rd edition, John Wiley and Sons,

India, 2004.

[3] William R. Derrick and Stanley L. Grossman, A First Course in Differential

Equations, Third Edition.

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