501 phys ِ prof. awatif ahmad hindi ُ enter reference 1- w.e boyce and r.c diprema,...

501 PHYS

Prof. Awatif Ahmad Hindi

�Enter

ReferenceReference1- W.E boyce and R.C Diprema , "elementary differential equations " 3rd edition (1975), johnwily2- E.A coddington , “ an introduction to ordinary differential equation “ , prentice –hali (1961) 3- E.Kreyszig “advanced engineering mathematics “ 7th edition , johnwily (1993)4- L.S. ross , “introduction to ordinary differential equations” 4th edition , john wily (1989)5- Abramowitz , M. stegun , I.A. hand book of mathematical function . dover, new York (1962)

ReferenceReference

6- hochstadt , H. “special function of mathematical physics “ hold , rineheart , winstone , new york (1961)7- Lebedev,N.N.Special Functions and their Applications,Prentice-Hall,Englewood Cliffs,N.J.(1965)8-Rainvile,E.D.”Special Functions,Macmillan,New York(1960).

Special Functions of mathematics

Integral Equation

Differential equation

contentscontents

Definition

Properties of the Beta and Gamma functions:

some examples

Special Functions of mathematics Gamma and Beta functions

DefinitionDefinition

We define the Gamma and Beta functions respectively by

0

11 1

0

( ) 0 (1)

( , ) (1 ) , , 0 (2)

t x t

x y

x e t dt x

B x y t t dt x y

Properties of the Beta and Gamma Properties of the Beta and Gamma functions:functions:

2 2 1

0

(1) 1 (1)

( 1) ( ) 0 (2)

( 1) ! (3)

int

( ) 2 (4)t x

x x x x

x x

x is a nonnegative eger

x e t dt

2

2 1 2 1

0

12

( ) ( )os sin (5)

2 ( )

( ) (6)

( ) ( )( , ) (7)

( )

( , ) ( , ) (8)

( 1, ) ( , ) ( )(9)

x y x yc d

x y

x

x yB x y

x y

B x y B y x

xB x y B x y i

x y

( , 1) ( , ) ( )(9)yy xB x y B x y ii

2 12 12

sin( )

(2 ) ( ) ( ) (10)

( ) (1 ) (11)

x

x

x

x x x

x x

Definition of the Gamma function for negative

values of the argument

Definition of the Gamma function for negative

values of the argument

12( ) ( 1)x x

22 2 2

2( ) 0 (1)

d y dyx x x y ydx dx

12( ) ( 1)x x

Some examplesSome examples

Express each of the following integrals in terms of Gamma or Beta functions and

simplify where possible

2

0

( ) tani d

32

12

1

3 30

0

1

1

)1

) (1 )

1) ( )

1

t

dxii

x

iii t e dt

xiv dx

x

Bessel FunctionsBessel Functions

Bessel’s equation of order n is

22 2 2

2( ) 0 (1)


22 2 2

2( ) 0 (1)


2

2 2 22

( ) 0 (1)d y dy

x x x y ydx dx

22 2 2

2( ) 0 (1)


We shall solve (1)by using Frobinous method the solution of (1) is given by :

2

0

2

0

1( ) ( 1) ( ) (2)

! ( 1) 2

1( ) ( 1) ( ) (3)

! ( 1) 2

rr n

nr

rr n

nr

xJ x

r n r

xJ x

r n r

The explicit relation ship between The explicit relation ship between andand

for integral n isfor integral n is

shown in the following theorems

Theorem 2

Theorem 6

Theorem 1

Theorem 3

Theorem 4

Theorem 5

Theorem 7

J

Theorem 1Theorem 1

When n is an integer (positive or negative) 2

2 2 22

( ) 0 (1)d y dy

x x x y ydx dx

22 2 2

2( ) 0 (1)


2

2 2 22

( ) 0 (1)d y dy

x x x y ydx dx

( ) ( 1) ( )nn nJ x J x

Theorem 2Theorem 2

The two independent solutions of Bessel’s

equation may be taken to be

22 2 2

2( ) 0 (1)


2

2 2 22

( ) 0 (1)d y dy

x x x y ydx dx

( )

cos( ) ( ) ( )( )

sin( )

n

n nn

J x

and

n J x J xY x

n

( )

cos( ) ( ) ( )( )

sin( )

n

n nn

J x

and

n J x J xY x

n

For all values of n.

Theorem 3Theorem 3

2

2 2 22

( ) 0 (1)d y dy

x x x y ydx dx

22 2 2

2( ) 0 (1)


-Generating function for the Bessel functions

1 12 ( ) 2 ( )tx t

nn

e t J x

1 12 ( ) 2 ( )tx t

nn

e t J x

1 12 ( ) 2 ( )tx t

nn

e t J x

Theorem 4Theorem 4

2

2 2 22

( ) 0 (1)d y dy

x x x y ydx dx

22 2 2

2( ) 0 (1)


1 12 ( ) 2 ( )tx t

nn

e t J x

1 12 ( ) 2 ( )tx t

nn

e t J x

Integral representations for Bessel functions:

1( ) cos(( ) sin ) ( int )nJ x n x d n eger

Theorem 5Theorem 5

2

2 2 22

( ) 0 (1)d y dy

x x x y ydx dx

22 2 2

2( ) 0 (1)


1 12 ( ) 2 ( )tx t

nn

e t J x

1 12 ( ) 2 ( )tx t

nn

e t J x

12

1122 1

2112

( )( ) (1 ) ( )

( )

nn ixt

n

xJ x t e dt n

n

2

2 2 22

( ) 0 (1)d y dy

x x x y ydx dx

22 2 2

2( ) 0 (1)


1 12 ( ) 2 ( )tx t

nn

e t J x

1 12 ( ) 2 ( )tx t

nn

e t J x

Recurrence Relations

1

1

'1

( ( ))) ( )

( ( ))) ( )

) ( ) ( ) ( )

nnn

n

nnn

n

nn n nx

d x J xi x J x

dx

d x J xii x J x

dx

iii J x J x J x

'1

' 11 12

21 1

) ( ) ( ) ( )

) ( ) { ( ) ( )}

) ( ) ( ) ( )

nn n nx

n n n

nn n nx

iv J x J x J x

v J x J x J x

vi J x J x J x

2

2 2 22

( ) 0 (1)d y dy

x x x y ydx dx

22 2 2

2( ) 0 (1)


1 12 ( ) 2 ( )tx t

nn

e t J x

1 12 ( ) 2 ( )tx t

nn

e t J x

graphs of the Bessel functions

Theorem 6Theorem 6

2

2 2 22

( ) 0 (1)d y dy

x x x y ydx dx

22 2 2

2( ) 0 (1)


1 12 ( ) 2 ( )tx t

nn

e t J x

1 12 ( ) 2 ( )tx t

nn

e t J x

Orthogonally of the Bessel functions

2 212

0

( ) ( ) { ( )}a

an i n j n i ijxJ x J x J x

If i jand are roots of the equation

( ) 0nJ a

Theorem 7Theorem 7

22 2 2

2( ) 0 (1)


1 12 ( ) 2 ( )tx t

nn

e t J x

1 12 ( ) 2 ( )tx t

nn

e t J x

If f(x) is defined in the region

-Bessel Series0 x a

and can be expanded in the form

1

( )i n ii

c J x

Where the i

are the roots of the equation ( ) 0,nJ a

then 0

2 21

2 ( ) ( )

{ ( )}

a

n i

in i

xf x J x dx

ca J a

2

2 2 22

( ) 0 (1)d y dy

x x x y ydx dx

22 2 2

2( ) 0 (1)


1 12 ( ) 2 ( )tx t

nn

e t J x

1 12 ( ) 2 ( )tx t

nn

e t J x

problems

1-Use the generating function to prove that

( ) ( ) ( )n r n rr

J x y J x J y

2- Show that

1

0

( )nn

J xdx

x

2

2 2 22

( ) 0 (1)d y dy

x x x y ydx dx

22 2 2

2( ) 0 (1)


1 12 ( ) 2 ( )tx t

nn

e t J x

1 12 ( ) 2 ( )tx t

nn

e t J x

Integral equation

Definition(1)

Defintion (2)

Defintion (3)

2

2 2 22

( ) 0 (1)d y dy

x x x y ydx dx

22 2 2

2( ) 0 (1)


1 12 ( ) 2 ( )tx t

nn

e t J x

1 12 ( ) 2 ( )tx t

nn

e t J x

An integral equation is an equation in which an unknown function

appears under one or more integral signs.Naturally, in such an equation there

can occur other terms as well .

Integral equation

Definition(1))

22 2 2

2( ) 0 (1)


1 12 ( ) 2 ( )tx t

nn

e t J x

1 12 ( ) 2 ( )tx t

nn

e t J x

For example

for

, ,a s b a t b for The equation

( ) ( , ) ( ) (1)b

a

f s k s t t dt

2

( ) ( ) ( , ) ( ) (2)

( ) ( , )[ ( )] (3)

b

a

b

a

s f s k s t t dt

s k s t t dt

Where the function

22 2 2

2( ) 0 (1)


1 12 ( ) 2 ( )tx t

nn

e t J x

1 12 ( ) 2 ( )tx t

nn

e t J x

for

( )s

is the unknown function which all the functions are known are integral equation .

These functions may be complex –valued functions of the variables S and t

2

2 2 22

( ) 0 (1)d y dy

x x x y ydx dx

22 2 2

2( ) 0 (1)


1 12 ( ) 2 ( )tx t

nn

e t J x

1 12 ( ) 2 ( )tx t

nn

e t J x

An integral equation is called linear if only linear operations are performed function on it upon the unknown function .

The equations (1) and (2) are Linear while (3) is nonlinear.

Integral equation

Definition(2))

22 2 2

2( ) 0 (1)


1 12 ( ) 2 ( )tx t

nn

e t J x

1 12 ( ) 2 ( )tx t

nn

e t J x

Integral equation

Definition(3))The most general type of linear integral equation is of the form:

( ) ( ) ( ) ( , ) ( ) (4)a

h s s f s k s t t dt Where the upper limit be either variable or fixed.

The functions are known functions, 1 1,f h and k

While

1 12 ( ) 2 ( )tx t

nn

e t J x

1 12 ( ) 2 ( )tx t

nn

e t J x

for

( )s is to be determined ;

is called the kernel .

is a nonzero,real or complex, parameter

( , )k s t Is called the Kernel

2

2 2 22

( ) 0 (1)d y dy

x x x y ydx dx

22 2 2

2( ) 0 (1)


1 12 ( ) 2 ( )tx t

nn

e t J x

1 12 ( ) 2 ( )tx t

nn

e t J x

Special casesThe following special cases of

equation (4) are of main interest :

I) Fredholm integral equation

II) Volterra Equations

I) Fredholm integral equation:

In all Fredholm integral equation of the first kind the upper limit of integration b,say,is fixed.

i) In the Fredholm integral equation of the first kind Thus,( ) 0.h s ( ) ( , ) ( )b

a

f s k s t t dt

( ) ( , ) ( )b

a

f s k s t t dt =0 (5)


ii) In the Fredholm integral equation of the second kind,

( ) 1;h s

( ) ( ) ( , ) ( ) (6)b

a

s f s k s t t dt


iii) The homogeneous Fredholm integral equation of the second kind is a special case of(ii) above . In this case ( ) 0f s

( ) ( , ) ( ) (7)b

a

s k s t t dt

II) Volterra Equations II) Volterra Equations

Volterra Equations of the first, homogeneous ,and second kinds are precisely as above except that

b s is the variable upper limit of integration.

Equation (4) itself is called an integral equation of the third kind

Singular Integral equation:Singular Integral equation:

Definition (4)

When one or both limits of integration become infinite or when the kernel becomes infinite at

one or more points within the range of integration ,the integral equation is

called Singular .

For example ,the Integral equationsFor example ,the Integral equations

0

( ) ( ) ( ) (8)

1( ) ( ) , 0 1 (9)

( )

s t

s

s f s e t dt

and

f s t dts t

Are singular integral equations.

Special Kinds of kernelSpecial Kinds of kernel

separable or degenerate kernels

Symmetric kernel

I) separable or degenerate kernels

A kernel to k(s,t) is called separable or degenerate if it can be expressed as the sum of a finite number of terms each of which is the product of a function s only and a function of only ; that is,

1

( , ) ( ) ( ) (10)n

i ii

k s t a s b t

II) Symmetric kernel

A complex-valued function K(s,t) is called

symmetric (or Hermitian) if *( , ) ( , ),k s t k t s

where the asterisk denotes the complex conjugate.For a real kernel, this coincides with definition

( , ) ( , )k s t k t s

Eigen values and eigen Eigen values and eigen functions functions

Eigen values and eigen Eigen values and eigen functions functions

If we write the homogeneous Fredholm equation as

We have the classical eigen value or characteristic value problem; is the eigen value and is the corresponding eigen function or characteristic function.

( )t

Relationship between linear differential equations and Volterra integral

equation:

Relationship between linear differential equations and Volterra integral

equation:

The solution of the linear differential equation

With continuous coefficients ( ) ( 1,2,3....., )ia x i n

given the initial conditions ' 1

0 1 1(0) , (0) ,........, (0) (12)nny c y c y c

may be reduced to a solution of some Volterra integral equation of the second kind

From this hypothesis and some mathematical treatment we reach to

)13()(),()()(0

dtttsksfsx

where

1 2

1 1 1 2 0 2

( , ) [ ( ) ( )( )]

( ) ( ) ( ) ( ) ( )

k s t a s a s s t

f s F s c a s c sa s c a s

Methods of solutionMethods of solution

We shall explain some methods for solving linear integral equations ;

These methods are :

1- Analytical methods2- Numerical methods

Analytical methods for solving Volterra integral equation:

Resolvent kernel of Volterra integral equation. The method of successive approximation.

using Laplace Transform.

Solution of integro- differential equations with the aid of the Laplace transformation.

( 1)n

x

( 1)n

t

in

in

( , )k x t

( 1)n

x

( 1)n

t

I)If the kernel is a polynomial of degree

in

i) If the kernel has the general form k(x,t).

ii) If the kernel is a polynomial of degree (n-1) in x or (n-1) in t.

iii) the kernel is dependent on the difference of the arguments.

in

Resolvent kernel of Volterra Resolvent kernel of Volterra integral equation integral equation

( , )k x t

( 1)n

x

( 1)n

t

in

in

In the three cases above we shall begin with Volterra integral equation of the form

And after some manipulation we shall have ( , ; )R x t

Where is called the resolvent kernel .( , ; )R x t

The method of successiveThe method of successive approximationapproximation

0 ( )x

( ),x

1( ).x

{ ( )}, 0,1,2,3,.....n x n

we got

, where

0 ( )x

( ),x

1( ).x

Suppose we have a Volterra type integral equation (14).Take some function Suppose we have a Volterra type integral equation (14).

Take some function continuous in [0,a]

into the right side of (14 ) in place of we got

Continuing the process, we obtain a sequence of

Functions where,

0 ( )x

( ),x 1( ).x

{ ( )}, 0,1,2,3,.....n x n

1

0

( ) ( ) ( , ) ( )x

n nx f x k x t t dt

Where the sequence ( , )k x t

( 1)n

x

( 1)n

t

in

in

{ ( )}n x converges as n

to the solution of the integral equation (14) ( )x

Using Laplace transformUsing Laplace transform

0 ( )x

( ),x

1( ).xwe got

, where

0 ( )x

( ),x

1( ).x

The Laplace transformation may be employed in the solution of systems of Volterra integral equations of the type

xs

i ijj=1 0

(x)=f (x)+ k (x-t) (t)dt ( 1,2,3,...., ) (15)i j i s

( ), ( )ij ik x t f xWhere are known

continuous functions having Laplace transforms .

0 ( )x

( ),x

1( ).xwe got

0 ( )x

( ),x

1( ).x

Taking the Laplace transform of both sides of (15) we get :

1

( ) ( ) ( ) ( ) ( 1,2,....., ) (16)s

i i ij Jj

p F p k p p i s

This is asymptotic of linear algebraic equations in

( )i p Solving it ,we find ( )i p

Analytical methods for Analytical methods for solving Fredholm integral equation:solving Fredholm integral equation:

The method of Fredholm Determinants

Integral Equation with degenerate kernels

( , )k x t

( 1)n

( 1)n

t

I)If the kernel is a polynomial of degree

in

The method of Fredholm Determinants The method of Fredholm Determinants

0 ( )x

( ),x

1( ).xwe got

0 ( )x

( ),x

1( ).x

The solution of the Fredholm equation of the second kind

is given by the formula

( ) ( , ) ( ) ( ) (17)b

a

x k x t t dt f x

( ) ( ) ( , ; ) ( ) (18)b

a

x f x R x t f t dt

Where the function

( , )k x t

( 1)n

x

( 1)n

t

in

in

is called

the Fredholm resolvent kernel of equation (17) and defined by the equation

( , ; )R x t

( , ; )( , ; )

( )

D x tR x t

D

Provided the Here,

are power series in :

( ) 0D ( , ; ) ( )D x t and D N

( , )k x t

( 1)n

x

( 1)n

t

in

in

1

1

( 1)( ) 1

!

( , ) ( , ) ( , ) ( , )

nn

nn

b

n n n

a

D cn

where

B x t c k x t n k x s B s t ds

1

0 0

( , )

1, ( , ) ( , )

b

n n

a

c B s s ds

c B x t k x t

Integral Equation with Integral Equation with degenerate kernelsdegenerate kernels

0 ( )x

( ),x

1( ).xwe got

0 ( )x

( ),x

1( ).x

The kernel 1

( , ) ( ) ( ) (20)n

k kk

k x t a x b t

The integral equation (17) with degenerate kernel (20) has the form

1

( ) [ ( ) ( )] ( ) ( ) (21)n

k kk

x a x b t t dt f x

( , )k x t

( 1)n

x

t

in

in

After some manipulation ,it has the form

( 1,2,...., ) (22)m km k mc a c f m n

Where

( ) ( ) ,

( ) ( ) ,

( ) ( )

b

km k m

a

b

m m

a

b

k k

a

a a t b t dt

f b t f t dt

c b t t dt

Numerical methods for Numerical methods for solving Volterra integral equation:solving Volterra integral equation:

using the trapezoidal rule

( 1)n

x

( 1)n

t

the trapezoidal rule the trapezoidal rule

0 ( )x

( ),x

1( ).xwe got

0 ( )x

( ),x

1( ).x

Consider the nonhomogeneous Volterra integral equation of the second kind

( ) ( , ) ( ) ( ) (23)x

a

x k x t t dt f x

To apply the trapezoidal rule , let and

( ), 1,nx a

t nn

Define applying the , 1(1)jt a j t j n

trapezoidal rule to the integral of (23) ,we obtain:1

1 10 02 2

1

( ) [ ( , ) ( ) ( , ) ( ) ( , ) ( )] ( ) (24)n

i i n ni

x t k x t t k x t t k x t t f x

0 ( )x

( ),x

1( ).xwe got

0 ( )x

( ),x

1( ).x

, 1,j n nt x j x x t the integration in (23) is over

, ,t a t x Thus for we take t x

( , ) 0 ,i j j ik x t for t x

the equation (24) can be written in the form :

0 0

11 1

0 02 21

( ) ( )

( ) ( ) [ ( , ) ( , ) ( , ) ], 1, 2,..., ; (25)n

i i i i j j i j j j ij

x f x

x f x t k x t k x t k x t i n t x

0 ( )x

( ),x

1( ).x

0 ( )x

( ),x

1( ).x

The system of equation in (25) can be written in a more compact form as

0 0

11 1

,0 02 21

[ ) ], 1, 2,..., ; ( , ), , (26)n

i i i i j j i j j ij i jj

f

f t k k k i n k k x t j i

After some manipulation , we obtain

10 112

0 1 22 2

1 0 0 0 0

(1 ) 0 0

(1 )

t

t tn n n nn

k tk

k tk tk k

0 0

1 1 (27)

n n

f

f

f

0 ( )x

( ),x

1( ).x

0 ( )x

( ),x

1( ).x

By solving the system (27) we find

0 1 2 2 1, , ,......, , , .n n n

Which is an approximatetion of the solution of (23)

Numerical methods for Numerical methods for solving Fredholmsolving Fredholm integral equation:integral equation:

of the second kindof the second kind

The approximate method that we will discuss here for solving Fredholm equation of the second kind:

( 1)n

x

( 1)n

t

( ) ( ) ( , ) ( ) (28)b

a

x f x k x t t dt

0 ( )x

( ),x

1( ).x

0 ( )x

( ),x

1( ).x

Are based on approximating the solution

of (28) by a partial sum:

1

( ) ( ) (29)N

N i ii

S x c x

( )x

On the internal (a,b).If we substitute from (29)

into (28) for there will be an error

Of N linearly independent functions 1 2 3 1, , ,....., ,n n

( )x

1 2 3 1( , , , ,....., , )N Nx c c c c c

0 ( )x

( ),x

1( ).x

0 ( )x

( ),x

1( ).x

x

Involved which depends on x and on the way

the coefficients are chosen 1 2 3 2 1, , ,......., , ,N N Nc c c c c c

1 2 3 1( ) ( ) ( , ) ( ) ( , , , ,....., , ) (30)b

N N N N

a

S x f x k x t S t dt x c c c c c

Our main goal is how we can find or impose N conditions of the approximate solution (30).

1 2 3 1, , ,....., ,N Nc c c c c

1 2 3 1, , ,....., ,N Nc c c c c

The Galerkin approximate method The Galerkin approximate method

0 ( )x

( ),x

1( ).xwe got

0 ( )x

( ),x

1( ).x

In this method the N conditions are established for the determination of the N coefficients in (29)

By making of (30) 1 2 3 1( , , , ,....., , )N Nx c c c c c

orthogonal to N given linearly independent Functions on the interval (a,b).

1 2 3( ), ( ), ( ),......., ( )Nx x x x

0 ( )x

( ),x

1( ).x

0 ( )x

( ),x

1( ).x

x1 2 3 1, , ,....., ,N Nc c c c c

We will use the definition of orthogonality on

1 2 3 1( , , , ,....., , )N Nx c c c c c in (30) , Then these N

conditions become

1 2 3 10 ( ) ( , , , ,....., , )b

j N N

a

x x c c c c c ds

0 ( )[ ( ) ( ) ( , ) ( ) ] ( 1,2,3,...., )b b

j N N

a a

x S x f x k x t S t dt ds j N

0 ( )x

( ),x

1( ).x

0 ( )x

( ),x

1( ).x

1 2 3 1, , ,....., ,N Nc c c c c

After some manipulation ,we obtain :

1 1

( ){ ( ) ( , ) ( ) } ( ) ( ) , 1,2,3,...., (31)b b bN N

j i i i i ji ia a a

x c x k x t c t dt ds x x dx j N

Differential EquationsDifferential EquationsDifferential EquationsDifferential Equations

1- Power series solutions about an ordinary pointthis will not be discussed here because it has been taken in the past {Bs.C}

2-Solutions for singular points; the method of Frobineous

3- Bessel’s equation and Bessel Functions

Series solutions of linear differential equations

The method of FrobineousThe method of Frobineous

0 ( )x

( ),x

1( ).xwe got

0 ( )x

( ),x

1( ).x

We consider the homogeneous linear differential equation

2

0 1 22( ) ( ) ( ) 0 (1)d y dy

a x a x a x ydx dx

And we assume that 0x

is singular point of (1) under certain conditions we are justified in assuming a solution of the form

0 00

( ) (2)r n

nn

y x x c x x

0 ( )x

( ),x

1( ).x

0 ( )x

( ),x

1( ).x

Where r is a certain (real or complex ) constsnt.Again,we write the differential equation (1)

in the equivalent normalized form2

1 22( ) ( ) 0 (3)

d y dyp x p x y

dx dx

where

Outline of the method of Frobenius:

1- Let a regular singular point of the differential Equation (1), seek solutions valid in some interval

0x

00 ,x x R

0 ( )x

( ),x

1( ).x

0 ( )x

( ),x

1( ).x

and assume a solution

0 00

( ) ( )r nn

n

y x x c x x

Where we write the solution in the 0 0c

00

( ) (4)n rn

n

y c x x

2- Assuming term by term differentiation of (4)Is valid ,we obtain

10

0

( ) ( ) (5)n rn

n

dyn r c x x

dx

0 ( )x

( ),x

1( ).x

0 ( )x

( ),x

1( ).x

and

Now we substitute the series (4),(5) and (6) for y and its derivatives, respectively ,into the differential equation (1)

3- Now we proceed to simplify the resulting expression So that it takes the form

22

020

( )( 1) ( ) (6)n rn

n

d yn r n r c x x

dx

10 0 1 0( ) ( ) .... 0 (7)r k r kk x x k x x

k

( 0,1,2,.....)ik i

r

nc

are functions of

Where K is a certain integer and the coefficients

( 0,1,2,.....)ik i are functions of r and certain of the coefficients of the solution of (4)nc

0 ( )x

( ),x

1( ).x

0 ( )x

( ),x

1( ).x

x

00 ,x x R

we must set

4- In order that (7) be valid for all X in the deleted interval we must set 00 ,x x R

0 1 2 ...... 0nk k k k

5- Upon equating to zero the coefficient of the

lowest power of , we obtain

a quadratic equation in r called the indicial equation of the differential equation (1) .The two roots of this quadratic equation in r , called the indicial equation of the differential equation (1)

0k

r k 0( )x x

0 ( )x

( ),x

1( ).x

0 ( )x

( ),x

1( ).x

x

00 ,x x R

The two roots of this quadratic equation in r , called the indicial equation of the differential equation (1).The two roots of this quadratic equation are often called the exponents of the differential equation (1) and are the only possible values for the constant r in the assumed solution (4) .Thus at this stage the unknown constant is determined .We denote the roots of the indicial equation by where Here denotes the real part of and of

course if is real ,then is simplify it self.

1 2r and r 1 2Re( ) Re( ).r r

Re( )jr ( 1,2)jr j

jr Re( )jr jr

0 ( )x

( ),x

1( ).x

0 ( )x

( ),x

1( ).x

x

00 ,x x R

6- Now we equateto zero the remaining coefficients in (7) ,we are thus led to

a set of conditions, involving the constants r which must be satisfied by the various coefficients in the series (4).

1 2 3 4, , , ......k k k k

nc

1r for r

nc

nc

1r r

1r

2r

1r

and

are real and unequal , then

is the larger root.

7- We now substitute the root into the conditions obtained is step 6 , and then choose the to satisfy these conditions . If the are so chosen , the resulting series (4) with

is a solution of the desired form. Note that if and are real and unequal , then is the larger root.

1r for r

nc nc

1r r

1r2r

0 ( )x

( ),x

1( ).x

0 ( )x

( ),x

1( ).x

x

00 ,x x R

1r for r

nc

nc

1r r

1r

2r

1r

and

are real and unequal , then

is the larger root.

8- if ,we may repeat the procedure of step (7) using the root instead of In this way a second solution of the desired form (4) may be obtained . Note that if and are real and unequal , then is the smaller root . However , in the case in which and real and unequal , the second solution of the desired form (4) obtained in this step may not be linearly independent of the solution obtained in step (7) . also , in the case which are real and equal , the solution obtained in the step is clearly identical with the one obtained in step (7)

1 2r r2r 1r

1r 2r

2r

2r1r

1 2r and r

I hope that the presentation

is useful

501 phys ِ prof. awatif ahmad hindi ُ enter reference 1- w.e boyce and r.c diprema,...

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