5-34

Exercise(s)
Theory Illustrative Examples

Chapter

Basic Concepts
-361

401 many branches of pure

ntegral equations form one of the most useful techniques in
process. It is one
analysis, such as the theories of functional analysis and Stochastic
on account of it
of the most important branches of mathematical analysis, particularly
importance in boundary value problems in the theory of ordinary and partial
differential equations. Integral equations occur in many fields of mechanics and
26
mathematical physics. They are also related with the problems in mechanical vibrations,
theory of analytic functions, orthogonal systems, theory of quadratic forms of infinitely
many variables. Integral equations arise in several problems of science and technology
and may be obtained directly froin physical problems; eg., radiation transfer problems
and neutron diffusion problem etc. They also arise as representation formulae for the
solution of differential equation. The differential equations can be replaced by an
integral equation with the help of initial and boundary conditions. As such, each
solution on the integral equation automatically satisfies these boundary conditions.

a number of years,
Integral equations have been encountered in mathematics for
can be
originally in the theory of Fourier integrals. One of the first results which
-eonnected to integral equations were the Fourier inversion formulae:

(x) = |2 u) cosEx d5 ..(1)

¢ (5) cos xdE ...(2)

It can be regarded that the relation (2) gives a solution of the integral equation (1),
a
where u() is an unknown function and or) is known function.
 Integral Equatjons

In 1826, another integral equation was obt.ined by Abel:

A
moves in
material point under the action of force of gravity
curve. It is required
the vertical plane (E,n) along some smooth
to define the curve such that material point, starting from rest
at a point P(r, ) of the curve with ordinate y, reaches the
p
point E,n) at any instant t. The absolute value of velocity Fig: 1.1

of the moving point is

p= /2gy-))
an--/2g(r- n)]sine
dt
Where is the angle of slope of the tangent with -axis.
dn
dt =

By integrating between the limits from 0 to

,
2g(y- n)sine
we have

m--/2p)f),
where o(n) = (sin@)-'F(y) =
=[-nro)án,F(a) 0,

known as the Abel's integral equation, where F(Y)is a given continuous function and ¢ (n)
is the unknown function. The time of descent can be computed, if the path is known.

The actual development of the theory of integral equation began only the end of Nineteenth
century due to the works of the Italian mathematician V. Volterra (1896), and principally to
the year 1900, in which the Swedish mathematician I. Fredholm published his work on the
method of solution for the Dirichlet problem.

Flal Integral Equation

An integral equation is an equation in which an unknown function, to be deternined,
appears under one or more integral signs. If the derivatives of the function are involved,it is
called an integro differential equation.

An integral equation is called linear if only linear operations are

performed in it upon the
known functions, i.e., the equation in which no non-linear functions of
the unknown
function are involved e.g..

O(r)= fr)+ ("K()o(5)dE ..(1)

(r) = [K(«)O\E)d ...(2)

is a linear integral equation.
Basic Concepts

and .(3)
is a non-linear integral equation

Here the function o(r) in the equation (1), (2) and (3) is the unknown function while all the
other functions are known. These functions may be complex value functions of the real
variables x and .

An equation of the form

a()(*) + F()+ ^[ K(x)(5)d =0

iscalled the linear integral equation, where (*) is the unknown function; a (x), F(*) and the
kernel of the integral equation K (x,E)are known function ; is a non-zero real or
complex
parameter, and the integration extends over the domain
2 of the auxiliary variable.
Integral equations, which are linear, involve the integral operator

L-[, K(*,)d
having the kernel K (,E). It satisfies the linearity condition
LGO,(E)+ Cz,(5)]= CL[O,(E)1+ CGL[O>(5))
where LLO)]=|, K(,2)($)dE and C,G are constants.
Linear integral equations are classified into two basic types:

I. Volterra Integral Equation: An integral equation is said to be a Volterra integral

equation if the upper limit of integration is a variable, e.g.,

a(r)0() = F(x)+ a[ K(r)0(5)4E

(i) When oa=0, the equation involves the unknown function , appears only under the
integral sign and nowhere else in the equation, then
F(*) =
^[K(«E)05)45.4> -,
is called the Volterra's integral equation of frst kind.
(ii) When a = 1, the equation involves the unknown function ¢, both inside as well as outside
the integral sign, then
(1) = F()+ [ K(«)9(3)a5
is called the Volterra's integral equations of second kind.
(iü) When c= l,F (*) =0, the equation reduced to
9(1)= ["K()0(E)d
is called the homogeneous Volterra's integral equation of second kind.
 Integral Equations

II, Fredholm Integral Equation: An integral equation is said to be Fredhol:n integral

e.g.,
cquation if the domain of integration Q is fixed,
a(x)0(r)= F(*)+ ["K(»)0(E)dG

() Whena a O, the cquation involves the unknown function þonly under the integral sign, then

is called the Fredholm integral equation of first kind.

(iü) When a l, the equation-involves the unknown function¢ both inside as/well as outside
the integral sign, then
o() =F(*) + a[K(x8)0(E)d, as xsb
is called the non-homogeneous Fredholm integral equation of seeond kind.
(iüi) When a = 1,F(x) = 0, the equation reduced to

called as the homogeneous Fredholm integral equation of second kind.

Linear integral equations of thefirst and second kind are the particular cases of the linear
integral equation of the third kind, (which are of the form I and I1), where a is not a
constant but is a prescribed function of x, i.e.,
= F(r)+
(x)() [, K(%)0(E)dE

III. Singular Integral Equation: When one or both limits of integration become infinite or
the kermel becomes infinite at one or more points within the range of integration,
the integral
equation is called singular integral equation, For example
+A
()= f*) exp{-x-E|}o(E)dE

and f)-f(-)"(E)AS, 0<a<l

are called the singular integral
equation. The second equation represents the Abel's integral
a
equations for =-1/2

IV. Non-linear Integral Equation: If the unknown

function appeats under an integral sign
to a power n(1> 1), then the equation is said to be a non-linear
integral equation. For example
(*) = F(*) + K(x,)O"(E)A

V. Convolution Ircegral: If the

kemel K(,) of the integral equation is defined as a
function of the difference (x-E), i.e.,
Bastc Concepts

K(») = K(*-).
where K is a certain function of one variable, then the integral equation
olr) = F()+ [K(x-E1(E)A ...(1)

and the corresponding fredholm equation

otx) = F(«)+ [K(r-)(E)d ...(2)

are called integral equation of the convolution type.

The function defined by the integral

..(3)

is called the Convolution or the Faltung of the two functions K and are known as the
¢ and

convolution integrals. In general, the convolution integrals may be defired as

K(*-)(E)d= K(E)O(E- Ž)A ...(4)

Integral Equation

Linear* Non-linear** Singulrt

Volterra Fredholm
(upper limit (domain of integration is fixed)
of integration
is a variable)

First kind Second kind Thitd kind

[ a(e) is a prescribed
F)= Kix.E)(E) d, functionof x]

Homogeneous
Non-homogeneous
Fig. 1.2

=
O(1) = F(r)+ ^, K(*5)(E)dE O(r) 4, K(:) (3)d
K(uE)9(;)4%,=0
t(x)(x)+ F()+ [,
**(x)(*) + F(r)+ a[,K()"G)d =0

= 0
a(x)o(r)+ F(x)+ [ exp.(-*-}(E)d
 Integral Equations

l.2 Différentiation of a Function Under an Integral Sign

Consider the function I,(r) defined by the relation

...(1)
where n is a positive integral and a is a constant.
We know that

Jr F(n)án -fF(n)) dn +
P F|xQ1)de-F[P))e,
which is valid if F and OF/ax are continuous functions of both x,n and the first derivative
of
P(r) and Qx) are continuous.
Differentiating (1 ) under the integral sign, we have
=
dr
(n-1)[x-n)f(nán + (*- n)=fn)hex
dl, = (1-1)/-],1> 1.
dx .(2)
From the relation (1), we lhave

..(3)
Differentiating (2) successively nn times, we have

=
"dm (n-1)(n-2) (n-3)....(n–m)ly-m,1> m
In particular, we have

= (1-1)!I (x)
dl-l
d
a=,=(u-)ln = (n-1)!f(r) ...(4)
dx

Thus, we have

dlp.

In general, we have

...5)
From the relations (1) and (5), we
have
Baslc Concepts

n-)u--nyfom)an
a x
This may be represented as the result of integrating the function f Irom to and then
integrating (n-1) times, we have

..(6)
(n-1)!

1.3 Relation Between Differential and Integral Equations

There is a fundamental relationship between integral cquations and ordinary and partil
differential equations with given initial values. Consider the differential cquation of nh
order as

d" d-I +..t a,r)y = F(r) ...(1)

with continuous coefficients a(x), i = 1,2,3,...The initial conditions are preseribed as

follows:
y0) = Coy'0) = Cy"0) =
Cz..ye\(0) = Gpel
..(2)
where the prime denotes differentiation with respect to x,

Consider -=(*)
By integrating and using the initial conditions (2), we have

oE)dE + Ch-l

-2Jo 5)d3° + C-*+ Cp-2

+G-2 ...(3)
(n-1)! (n-2)!;t.+G

($)d"represents for a
n.
where multiple integral of order

From the relation (3) and (1), we obtain

=
F(x)+ iCa*) ...(4)
 Integral Equatlons

where ..(5)
(n-i)!
-10(5)dE
(n-1)!
=G(1) ..(6)
where G(r) = F()+ SiCx,(*)

The equation (6) represents the non-homogeneous Volterra's integral equation of second
kind.

Particular Case
Consider the linea differential equation of second order.
+
a(r)+a, (a)y = F(a) ...(.1)
d
with initial conditions
= and y'(0) = G
J(0) G ...(2)

Consider "=()
By integrating and using the initial conditions (2), we have

dx ..(3)

and y= +
(x-E)o(Š)d5 Cx+ G ..(4)
The given differential equation reduces to

o(x)+

or o(r) + [ La(x)+ a,(x)(r-)10(5)d5

=F(x)-Ca, (x)-C;xa,(x)-Coag(x)
or 9(x)=f(r) +2[ K(xž)óE)dF
..5)
Where K(,) = a (x)+a, (x)
(r-),=-1
f(x) = F(x)- Ci4,(x)-Cixa,(x)-Gag (x)
..(6)
which represents the Volterra's intègral
equation of the second kind. Similarly,
value problems in ordinary differential the boundary
equations lead to Fredholm integral equatidns.
Baslc Concepts

Ilustrative Examples
Evmple l: Show that the function ¢(x)= xe* is a solution of the Volterra integral equation

o(*) = sinx+ 2[ cos(x-0(Š)dE

[Kanpur 2005, Meerut 2009]
Solution: Substituting the function o(*) = xe
in the given integral equation, we have
=
sinr+ 2(, coc(r-E).e dk

-sinr+2| cos cosk až +

sinr,e sinšd,|

= sin
+2| cos x
x

+sin
ing- cosi)
lo
-sing- cor)4;|

-sinx+2 coscosr+ sinn

(cosk + sin + sin- cos)

osri
-sine oin - cosk- cos-sin)
= sin .x +
e' (cosx+sin x)- cos x(e sin E), +
sin x(eš cos )
= sinx+ xe- cosxesinx)+ sinx(e* cosx-1).

= xe= L.H.S
Hence (x) = xe* is a solution of the given integral equation.

Evmple 2 Show that the function ¢ (x) =(1l+ y

is a solution of the volterra integral
equation

o)=Jo )
[Kanpur 2009]

ax
cos b cos bx
=[a
bx dx cos br +
bsin bxl|Sin
bx22 -[asin br
 10 Integral Equations

Solutjon: Substituting the function (x) (|+

= x
in the given cquation, we have

1+
1+\1+
+
1+ (l+2 1+2 (1
+22
x, thus,
The substitution of (x) reduces the given equation to an identity with respect to
(*) =(1+23 is a solution of the integral equation.

Exaniple 3: Show that the function o(x) = 1/TVx is a solution of the integral equation
p(3)
Jo
Jr-t) dE =1
Solution: Substituting the function o(x) =1/Tx in the given equation, we have

=dE

Thus o()= 1/Vx is a solution of the integral equation.

Example 4 Showthat the function (r)=1-xis a solution of the integral equation

o(E)dE= x

Solution: Substituting thefunction ¢

(r)=l-xin the given cquation, we have

= e*(l-e*)+ e(e+e-l)=x= R.H.S.

Thus o(x) =1-x is a sòlution of the integral equation

Baslc Concepts

Exmple 5: Show that the function (x) = is a

l solution of the fredholm integral equation,
o(*)+ [ tes -I)($)d =
e"-x
Solution: Substituting the function
¢(x) = l in the L.H.S. of the equation, we
have

=1+(e*-x-l)e-x=R.H.S.
Thus o(x) = l is a solution
of the given integral equation.

Exanple 6:
Show that the function o(a)=
e[2x- (2 /3)1 is a solution of the fredholm
integral cquation.

9()+ 2[Jo eso(š)AE = 2xe",= 2

Solution: Substituting the function
o(x)=e[2x-(2/3)] in the LH.S. of the equation, we
have

-(2r-)•2r-)
-2xe = R.H.S.

Thus o(x)= e*|2r-| is a solution of the given integral equation

3

Example 7: Show that o(x)= cos2x is a solution of the equation

o(x) = cosr+ 3 K
(x,)(E) d
x cos E,0<xsE
where K(x,)= sin
cosxsin ésSn
[Meerut 2004, 2008; Kanpur 2005]

Sol: o(x) = cosr+3 [ K(x)(5)d5

t>0
o
(x) = cos x +3 cos x
sino () dz t-0 t =

+[ sinrcosšo(E)A | ..(1)

Since o(x) =
cos2x is a solution of the equation, so it must satisfy the equation (1).

Hence o)= cosr+ 3 cos x

sinžcos 2 dE + sin xcosskcos 2%,45
 12 Integral Lquations

x (cos 35 + cos
, )d
or +cos
¢ () = cos x + [ (sin 3 -sin E) d + sin
T

o (x) = cos x + 3
5-cos cos35– cos &) + sin
sin 35 +sin &)|
o(r) = cos*+cos2r-cosx
or o(r) = cos21,
shows that it is a solution of the given integral equations

Bxample 8: Show that the function o(u)

=
sin(u/2 ) is a solution of the equation

x2-),0sxsk
where K()= f
5(2-x),sxsl
[Meerut 20051
K(».E)0(3)dk
Solution: Here o(1)=r*
+
o)=r*J$0-9sin f:2-B) sin
-ž+2-)Esin 5+* 2-)sin
2

2 Jo

4 sin

Z(2-x)cos- 4 sin
2

x)xcos (2-x)sin
2 2

+x2-x) cos 2 2 in TUX

-rsin=sin T
TUY

Hence the function o(x) = sin(Tr/2) is a solution of the given integral equation.
Basic Concepts 13

Erample 9:
Verify that the given function o(x) = is the solution of the integral
(l+2
cquation,

(1) =. 3.*+2,3 ra3r+2*-o(E)d

3(1 + (l+2
Solution: Substituting the functions o(x) =
in the given equation, we have
(1+2
3x+ 2,3 -x3x+ 2x
3(1+2 (1+2
3*+2r3 x
3x+2r3
3(1+ ((+ (+852 l+572
3x+2x3 3x+2x3 1

30+27*3q+ la+gg7*30+2a+
3x+2x3 3x+23
+
3(1+2
3x+2y3
=L.H.S..
(l+2 |3(1+ 3(1+*|

is a solution of the given integral equation.

=
Thus, o(r)
(+F2

Evample T0: Verify that the given function o(r)

=is the solution ofthe integral equation.

Solution: Substituting the function (x) = in the given equation, we have

¢
which is obvious.Hence, the function (x)=satisfies the integral equation.

=sin" cos 9d =sin = 1E3

1

tan" secde;¿= tan e, d = sec 9de (1+¿2972

0
(1+E2 sec e
33
 Integral Equatlons

E XERCISE-1.1

1. Verify whether the given function o(r) are solution of the corresponding Volterra's
integral cquation.

(a) o() =*-r:0(*)

6 =x-,sinh(x-)o(5)d5twnt
(b) o(*) =e(cose*-e'sine);

e(r) = (1- xei)cosl-' sinl + [ l-(*-)e"} oE) dE

(c) o(1) =
3;-ko($)Jd5
(d) o(r) =1-*[;oE)d =x

2. Verify whether the given functions, ¢(*) are solutions of the corresponding Fredholm's
integral equations:

(a) 9(7) = c'so(*) = 1- [, sinažO(E)dE

(b) (*) = x*"o(r) =(*-)e*+4[+5)o(E)aE

(c) 9(x) = cosx;Q(x) = sinx + [ (+) cosO(5)d

(d) o(x) =1 2sinx =

1+ cos(r+)O(E)AF
1-(r/2)
:o()

Evample 1Ua Form an integral equation corresponding to the differential equation

y
dy
-5 +6y =0
d
with the initial conditions
y(0) =0,y (0) =-1
Solution: Consider =(*) ..(1)
d
By integrating with regard to x, we have

Jo E) dž + A = J, ) dž -l, since y'(0) =-l ...(2)

 Basic Cóncepts 15

and y= f, oE)Ag -x+B, since y (0) = 0

Or ...(3)

Substituting the relations (1l), (2) and (3) in the given differential equátion, we
obtain

Or ) = (6x-5)+ ( (5- 6x+ 6)6)d

represents a Volterra's integral equation second
of kind.
Example 12:| Forn an integral equation corresponding to
the differential equation
y

dx
2+y=0
with the initial conditions
y0) = 1,y'(0)= 0
y
Solution: Consider = r)
...(1)

Then dy +A= [
-(Š)d E)aE, since y'(0) = 0 .(2)
and y=Jo (*-)(E)A + B

= 1
Or
J= J, (*-E)(E)d +l, since y(0) ...(3)
From the relations (1), (2) and (3), the given differential
equation reduces to

ox) = -1-[, (2x-E)0(5)dš,

represents a Volterra's integral equations of second kind.

Example 13: Form an integral equation corresponding to the differential equation.

y -sin+ e*y=x
d? dx

with the initial conditions

=
y(0) =ly'(0) -1
[Meerut 2002, 2004]

Solution: Consider ..(1)

d
 16 Integra Equattons

Then
..2)
and ..(3)

we have
Substituting the relations (1), (2) and (3) in the given differential cquation,

(r) = [*-sinx+ e(x-1)]+ x-e(x-E)]o(JAE

[ Isin

represent a Volterra's integral equation of second kind.

Examplé 14: Fornn an integral equation corresponding to the differential equation

-+ (-)y = xe*+1
d3
with initial conditions
y(0) =l= y'(0), y"(0)= 0.
y
Solution: Consider .=(*).

Then
y .(1)

- -E)E) d + 1

dx

and y= (- ¿' o(E)d + x+l ...2)

Substituting the relations(1) and (2) in the given differential equation, we have

o(x)+

=
¢(*) xe* +1- x(r-)-[+t-)(r-E)']o() dk

represents a Volterra's integral equation of second kind.

Exaimple 15: Form the integral equations corresponding to the following differential
equation with the prescribed initial conditions:
(i) +y = cosx; y(0 ), y'(0)=1
d2
(i1) -2:1) = 0;
y(0) =y0)=] = y'0)
Basic
bncepts 17

y
Solution: Consider = (1)
d
Then - E)A +1,

and y= (x-E)(4)dE+ x
Substitting the relations (1) and (2) in the given differential
equation, we have
(x)+ [(*-)(E)d5+ x= cos*
Or o(*)= (cosx-*)-
f (x-E)(E)AG
represents a Volterra's integral
equation of second kind.
y
(ii) Consider = (x)
dy ..(1)

Then dy =
[oE)d5 +1
dy - (-)O()A +
x+1

and y+ (-&o(E)A+}+x+} ...(2)

Substituting the equations (1)and (2) into the given differntial
equation, we have
)-2 r-toE)45++x*-0.
9(*)= x(*+ 1 + [xr-o(E)d
represents a Volterra's integral equation of second kind.

Example 16: Convert thedifferential equation

y"(x)-3y'(r)+ 2y(r) = 4sinx,

with the initial conditions y(0)= 1, y'(0)=-2 into Volterra's LE. of second kind.
[Meerut 2003]
=
Solution: Consider
d
(*)
dy

J= (x-E)()d-2x+1
o)-3|oGMG-2|+2 -9«E)E-2*+1|=dsinx
 18 Integral Equalona

o(r) =
4(-2+ r+ sinr) + [3-2(x-E)19(3)d%
=
4(-2+ x+ sinr),
=
where F(x) 1

and K(r.) = 3-2(r-E)

which represents the Volterra's integral equation of second kind.

EvampleI7: Reduce the initial value problem

"(r)+ A ¢(*) = F(x)

with o(0) = 1,¢(0)=0

SolutioT: The differential equation is given as

"(r)+ 2o(x) = F(*)

"(*)=F(*)- Ao(r)

Integrating both the sides with regard to x, we have

¢(*) = F()- 0(*) }dx

Integrating both the sides with regard. to x, we have

o(x)- 0)=F(G)– A9(E)}a?

o(:) = 1+ [(*-)F(E)- 0(E)} d5

which reduces to a Volterra's integral equation of second kind.

Baslc Concepts 19

EXERCISE-1.2
"Form the integral equations corresponding to
the following differential equations with
the given initial condition
dy
1. (a)
dx -J=0; y(0) =1

(b)
dy
ty=;y (0) =1, y
(0) =0
[Meerut 2009]
2. -6+5y
dx =0;y"(0) =G,y (0) = Cz,y(0) =C3ht
3.
d2 -+y=cos
y
*; y(0) =0= (O)

y
4. +(l+)y= cos x; y(0) =0, y
(0) = 2

dy y
5. +6y =0; (0) =0, y
(0) =1

6. Show that a linear differntial equation with constant coefficients

reduces, under any
initial conditions, to a Volterra's integral equation of the second kind.
dy
7. --3y = 0; y(0) = 1 y (0) =0
dx
dy +yy
8. =;y (0) = y (O) =0

A)NSWERS-1.2 r

o(x)=G(3 + 6x-(5/2)**)+ (6-5x)-5Cg +[[3+6(x-E)-(5/2)(*-05)dk

O(*) = cos *-S (-)¢ (E)
E
o(1) = cos x-2x(l + *)- + *-) ($) d5

(*) =5-6x+ [5-6(x-E)]4(Š) dE

 20
Integral Equations

1lustrative Examples

Evnple18 Reduce the initial value problem

e =x
'"- sinxo'+
with the conditions o(0)= 1,o(0)= -1
Conversely, derive the
to a non-homogeneous Volterra's integral equation of the second kind.
integral equation obtained.
original differential equation with the initial conditions from the [Meerut 2004, 2007)
as
Solution: The given differential equation may be written
= *-eo(r)+ sinxo'(r)
"()
Integrating with regard to x, both the sides, we have

S o)dr =o xá-f,eo(*)dr f sinx(r)de

+

cosxo(1)dt
[o; --f6e'o)d*+[sinxo(r)6-fi
(r)-¢0)==+ sinx¢(*)- [ (e* +
cosx)¢(*)d*
2
= sinxo(x)- [(e* +
cosr)¢(r)dr
(r) -1+-+
2
Integrating both the sides with regard to
x, we have

-l+*)áx + [ sinro(r)du
Sdu=
(e*+ cos.x) O(z)d?
-I
9(1)- (0)=-*+-+ sin o()d
6

-Ii (*-E)e5 + cos)Q(E)d

-
o(2)=(1-x+)+[sin
6
(*-)(eš + cos)J0(E) dG ..(1)

represents the non-homogeneous Volterra's integral equation of second kind.

Converse. Again,differentiating the integral equation (1) with regard to x, we have

o() =-1+ sinž-(r-E)(eŠ + cos)]0() dG

2
-x +
O() =-1+;+[[sinž- (*-)(e cos)]J0(5)d
+[sinx- (r-r)(e* + cosx)]o(*)1

(r)=-1+.
2
2-[(e+ cos)()dE+ sinxo(r) ...(2)
Basic Concepts 21

Differentiating with regard to x, both the sides, we have

$"(1) = - + cos)o(E)A + cOsx¢(*)+ sinx¢(x)

o"()= + cos)0«E)
x-(e A + (e* + cos:)9\)1
+ cos xo(r)+ sinx¢'(x)
(r) = *-(e*+ cos x) o(*) + cos xo(x) + sinx¢(*)
"(x)- sinr¢(r) + e"%r)=x
which is the required given differential equation. Putting x
we have
=0 in the equation (1) and (2),

(0) =land '(0)=-l

Exanple 193 Reduce the differential equation

¢"(*)-3¢(r)+ 20(r)=4sinx
with the conditions
(0)= 1,¢ (0) = -2
into a non-homogeneous Volterra's integral equation of second kind. Conversely,
derive the'
original differential equation with the initial conditions from the integral
equation obtained.
[Meerut 2003]
Solution: The given differential equation may be written as
"(x)= 4sinx- 2 0(r)+ 3¢(*) ..(1)
Integrating with regard to x, both the sides, we have
i )dr =4f, sinxdx -2[ o(*)dr +3[(*)d*

[)K=-4(cos ), -25 ()de + 30()H

()-o(0)=-4(cosx-1)-2[, o(«)de +3(0(*)- I)
+
o(*) =-2-4(cos*-1) 3[9(*)-1]-2, o(x)dk
(*)=-1-4 cosx+3(x)-2 (x)dx i..(2)

¢(*)=-1-4 cos x + 3(r)-2 *dr

Integrating with regard to x, both the sides, we have
dr -4{, cosxdr + 3
[, rdr -2 [ o(rdr?
S ode =-[
[o(*)6 =-x-4sinx+3 (r)dr -2 ( r)d

Q(x)- o(0)= -x-4sinx+3 [ (5)dž -2 [ o(G)E

 Antegral quations

9(r)-1=-- 4sinr+ 3 5)d5-2( (*-E)0(E)45

9(x) = (1-x- 4sinz) +

3-2(*-E)]0(E)dE ..3)
represents the non-homogernous Volterra's integral equation of second kind.

with regard to x, we have

Converse. Again, differentiating the integral equation (I)
(x) =-l–4 cosx+ 3-2- E)]0($)45
x)]0(r).l
¢(r) = -1-4 cosr+ [3-2(r-)lo(5)dE +[3 -2(x-
(r)=-1-4cosx+ 30(x)-2( o(Š)d5 ...(4)

Differentiating both the sides with regard to

x, we have

¢"(*) = 4sinz+ 30(*)-2 oŠ)d

"(r) = 4sinx 3¢(r)-2 o(r)
+

"(x)-3(x)+ 20(x) = 4sinz

which is the required given differential equation. Putting x =0 in the equations (3) and (4),
we have
(0)= 1,o'(0) = 2
Evnmple 20: Convert the differential equation

-3,y= 0
d
with initial condition y(0) =l,p'(0) =0 to an integralequation
Solution: Consider =o(*) ...(1)

Integrating with regard tox both the sides, we have

as y'(0) = 0
Y-[(E)AE,
dx
...(2)

Integrating again with regard to x both the sides, we'have

y()- y(0) = 5)d as y(0) =1

y(r) =l+[ -E)(E)d ..(3)

Substituting the relations (1). (2) and (3) in the given differential equation, we have
 Basic Concepts 23

9(*) =3+ f,(Sx-3)0(E)45

represents a non-homogeneous Volterra's integral equation of second kind.

Evmple 21:| For what value of the function o(r) = 1+ Ar is a solution of the integral cquation

Soution: The integral equation is given as

..(1)
The function o(r)=l+Ax satisfies the integral equation (1), then

*=fi(l+ a)d

r=-ele+hxe4 he-1-2]
x=-1+r+-e*-e]
+
x=-[1+ A(r+1)- e*(2 1)]

x+1+ 2(r+ 1)-e*(A+ 1) =0

x( +
1)+ I(2 +1)-e(a+ 1) =0

(a+ 1)(*+1- e*)= 0

=-1
Thus, the function (x) =1+ris a solution of the integral equation for h=-l
Evample 22: Convert the differential equation

-2x3¢=
dx
0

with initial conditions o(0) =0,¢(o) =0 to Volterra's integral equation of second kind.
Conversely, derive the original differential equation with the initial condition from the
integralequation obtained.
Solution: The given differential equation may be written as
„d = 30
d?
2x+
dx
we have
Integrating with regard to x, both sides,
24 Integral quation

do )
d +3 Qd
do

do

do =
2x0(r)+ Qdr

Integrating with regard to x, both the sides, we have

(05=2fo xo(r)dr +

+
o()= 2[, 5o(5)d5 [((-E)O(E)dE
..(1)
which represents the Volterra's integral equation of second kind.
Converse. Again, Differentiating the integral equation(1) with regard to x, we have
O(x)= (+E)E)d
E)oCE)dE
()-[6(*+
(*)=(3)d5+ 2xox
..(2)
Differentiating (2) with regard to x, we have
=
o"(x) E)AE+ 2(x$(*)+ o(x)}
dy

"(*) = ¢(*) + 2x¢(x) + 2o() = 0

'"(x)–2x(r)-36(r) =0
which is the given differential equation
From the relation (1) and (2), we have
o(0)=0 and '(0 ) =0
Example 23 Convert the differential equation

-+ ho = 0
d
with initial conditions (0)=0,¢(l) = 0., into Fredholm integral
¢
equation of second kind.
Also, recover the original differential equation from
the integral equation you obtain.
[Meerut 2002]
Solution: The differential equation may be
written as
Baslc Concepts 25

..(1)
d
Integrating both the sides with regard to x, we have

JO 2 ar=-(Qdt

$'(*)- (0)=- [ r)dr

Consider '(0) = C, a constant, then
' (r) =C-(*)dx
Integrating both the sides, we have

o(«)- Q(0)=Cr- [ (-E)(E)dt

o(*)= Cr-[ x-)(E)AE ...(2)
Since o() =0 /) =
Cl- [ ((-E)($)dE

1
C--E)oE)A ...(3)
From the relations (2) and (3), we have

o(*) = *fo (-E)9(E)A5-|;(-)0()d

(*) = 2( K(r)0(5)dE, ...(4)

E(-x) if 0
<Sx
where K(x,ž) = -2,ifx sžsl ..(5)
 Integral Equatlons

kind.
Fredholm integral equation of second
This detemnes the homogcneous
'The integral cquation
may be taken as
Converse.
O()= -(E)A-[a(r-E)o(E)A .(6)

x, we have
Differentiating both the sides with regard to

()-Jo

x, we have
Differentiating both the sides with regard to

drJo

o"(*)+ 2 o(r) = 0
From the relation (6), we have
o 0)
=0 and o0-f, -0-)ocE)d =0 .7)
which is the required given differential equation.

Example 24: Obtain Fredholm integral equation of second kind corresponding to the
boundary value problem
o=1-xo(*)
d
Also, recover the boundary value problem from the integral eq" youobtain

Solution: The given differential eq mnay be written as =1-xO(*). Integrating both the

sides with regard to x, we have

do
dx d

Consider '(0) = C, then

d-C+*-(xo()dr
d
 Basic Concepts
27
Consider '(0)= C, then
do =C+ x- [ xo(r)dr
dr
d -C+x-ix(*)ldr
Integrating both the sides with regard x, we
to have
(06= (c+ x)år- (rpd
o(:)- o0) = Cr+}-[ 50(5)dE

o()= Cr+--)50(E)A5
Since (l) =1

C-}+fa-)BO(E)A
o(*)=x(1+ *)+[, x1-)0E) A-I (-)o(E)
(*) =}x(l+ *) + , x(l-E)OE) E- -E)5(E) dE,
d5

+[x1-)%o(E)dE
«) =r(l +t x)+[l-x)$ oE)dš +[4-)*0(5)d,
o() =}l+ x)+ [ K(:5)OCE)A
.(1)

where K(4) x).<*

=-ažxx ...(2)

This determines the non-homogeneous Fredholm integral cquation of second kind.

Converse. The integral equation may be taken as

..(3)

Differentiating both the sides with regard to x, we have

d cl d
()=}+2x)+ -)5O(E)AF --)3O(E)A
d

–r-))
O()=1+21)+ftl-E))O(G)AF o(E)A

o()=}1+2x)+ [ -E)3O(E)AF-[žo(E)A .(4)

 e

Integral quations
28

x, we have
Differentiating again both the sides with regard to

¢(a)
=
1+ -)50(5)d–
o"(*) =1+ [-E)) O(E)d -[i 5} O(E)d- xo()
JO ax

"(*) =1- xo(r)

+
"() xo(1) =1 ...(5)

From the equation (3), we have o(0 )=0 and o(1) =

1,
...(6)

which is required differential equation together with the boundary conditions (6).
Evample 25: Obtain Fredholm integral equation of second kind corresponding to the
boundary valuc problems

+ ao = ;¢(0)=0,¢(|) =0

Also, recover the boundary value problem from the integral equation you obtain.

Solution: The differential equation may be written as [Meerut 2000, 2005]

d
Integrating, both the sides, with regard to x, we have

o(1)- o(0)=}-[o(E)d5

Let o'(0 )= C, then, we have

=
) C+}-E)d5
Integrating both the sides with regard to x, we have

(r)- ¢(0) = Cr+ (|/6)- (E)ag?

o(r)= Cr+ (/6)* -[ o5)d?

o(r)= Cr+ (/6)* -[ -E)O(5) dz
 29
Basic Concepts

=
Since o) =0 ) C+}- [o(5)5

C=-}+f, o5)45

o)=-/2) x+4/6)P + *f (9) d5 + f, xoG)4}

o)=4 /6)(-* -3a)+aB03)B+ f|xeE)AE

o(x)=(1/6)(P-3a)+ 2f; K(r)0(E)d ...(1)

where K(x.) =
En
This Determined the non-homogencous Fredholm integral equation of second kind.

Converse. The integral equation may be taken as

o(x) = -(1/2)x+ (1/6)+[ x(G) dE

-[6 (*-B)o(E)ds ...2)

Diferentiating (2) with regard to x, both the sides, we have

o()=-1/2) + /2)²+ (4
–-)o()d
axo(E)45
O)= - (1/2)+ (1/2) +
aj, o($)45 - [; E)a% ..(3)

Differentiating again with regard to x, both the sides, we have

d rx
o"a) =x+45)45– E)d

"(x)= *- Ao(x)
¢() +à (x) = x ..(4)
 Integral Equaatlone

From the relations (2) and (3), we have

o0) =0 and o)=-;+=0,

2
..(5)

conditions.
equation together with the boundary
which the required differential