Integral Equation

7/23/2019 Integral Equation

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1. Methods of Solution

Fredholms IE

In this section, we will discuss some methods of solution to Integral Equations that is Exact solu-

tion in relation to their existence sole dependence on the nature of the Kernel and Approximate

solutions in terms of the Free term and the Kernel.

Exact Solution

An exact solution is possible is based on the separability of the Kernel. That is if the Kernel can

be written ask(x, t) =g(x)h(t)for an IE

u(x) = f(x) + b

ak(x, t)u(t)dt (2.1)

then the IE is expressed as

u(x) = f(x) + b

ag(x)h(t)u(t)dt

whereg andh are functions ofx andtonlyrespectively. Then the Fredholms equation may be

solved as follows:

Since g(x) is independent oft, the integration with respect to t treats g(x) as a constant.Therefore

u(x) = f(x) +g(x)

ba

h(t)u(t)dt

Let

c=

ba

h(t)u(t)dt (2.2)

The solutionu can then be expressed as

u(x) = f(x) +g(x)c (2.3)


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2 Chapter 1. Methods of Solution

To solve for c we use yet the identity of the solution by changing the variable x to t. That is

u(t) = f(t) +g(t)cby substituting it into(2.2).

c = b

a

h(t) (f(t) +g(t)c)dt

c = b

ah(t)f(t)dt+c

ba

g(t)dt

c

1

ba

h(t)g(t)dt

=

ba

h(t)f(t)dt

c =

ba h(t)f(t)dt

1 b

a h(t)g(t)dt

The solution ofc is then substituted into(2.3).

Let us consider the example below:

Example 2.1

Solve completely

y(x) = 1

0xty(t)dt+x2

Solution

We first ascertain whether the Kernel can be expressed as a product of two functions in the form

k(x, t) =g(x)h(t). This is possible becauseg(x) =xand h(t) = t. Therefore

y(x) = x

10

ty(t)dt+x2

let c =

10

ty(t)dt

y(x) = xc +x2

By changing the variablesxto t, we havey(t) = tc +t2. We substitute this into the equationinvolvingc. Therefore

c = 1

0

t(tc + t2)dt

c = 1

3c +

1

4

c = 3

16

y(x) =x2 3

16x

Example 2.2

Solve completely

u(x) =coshx 12

x +13

10

xyu(y)dy


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3

Solution

u(x) = coshx 1

2x +

1

3 1

0xyu(y)dy

= coshx 1

2+

1

3x

10

yu(y)dy

= coshx 1

2x +

1

3xc

where

c=

10

yu(y)dy

c = 1

0y

coshy

1

2y +

1

3yc

dy

1 1

9

c =

10

y coshydy 1

6

10

y coshydy = y sinhy|10 1

0sinhydy

= e1 e1

2 e

1 + e1

2 1

= e1 + 1

Therefore

c= 9

8e1 +

5

6

u(x) = coshx

x

2+

1

315

16

9e1

8

x

= coshx 3

16

1 + 2e1

Example 2.3

For the homogeneous Fredholms

u(x) =

2

0(sinx)(sin t)u(y)dy

there is a solution for a particular value offor which the solution is non-trivial. Find this (eigen value) and the corresponding solution for u (the eigen function)


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Solution

We write the above as

u(x) =sinx

2

0sinyu(y)dt= csinx

wherec=

2

0 sinyu(y)dtBy changing variables and solving for c we have

c=c

2

0sin2ydy

Ifc =0, then= 4. The solution corresponding to this value of is

u(x) =A sinx

whereA is a constant.

Approximate Methods: Successive Approximation

USING THE FREE TERM

Let the Fredholms IE of the second type be given by

y(x) = f(x) + b

ak(x, t)y(t)dt (2.1)

We solve this Integral Equation on assumption that the Kernel of the equation k(x, t) iscontinuousin the square R : a x b, a t b and the function f(x) iscontinuous in theinterval[a, b]. These conditions will ensure thatk(x, t)and f(x)are bounded. We now look forour solution in a form of series of an ascending order of

y(x) =0(x) +1(x) +2

2(x) + +n

n(x) + (2.2)If the series above uniformly converges for some value of, then it can be substituted into the

right hand side of (2.1), replacing the argument x by tand effecting the term-wise integration

y(x) = f(x) +

ba

k(x, t)0(t) +1(t) +22(t) + +

nn(t) + dt

y(x) = f(x) +

ba

k(x, t)0(t) +2

ba

k(x, t)1(t) + +n+1

ba

k(x, t)n(t) +

Replacingy(x)by the approximate series. By equating coefficients of equal powers of, we

have

0(x) = f(x)

1(x) =

ba

k(x, t)0(t)dt

2(x) =

ba

k(x, t)1(t)dt

.......... .. ...............

n(x) =

ba

k(x, t)n1(t)dt (2.3)

The process of constructing the functionn(x)is called themethod of successive approxi-mationof solutions, and can be continued indefinitely.

The set of expressions helps us to evaluate the coefficients of the series (2.2) successivelyand to form the series, which formally satisfies the IE (2.1).


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5

Justification of the Method of Successive Approximation

For the sum of the series (2.2) to be a solution of the IE (2.1), it is necessary that it converges

uniformly.

Indeed suppose the kernel,k(x, t)is bounded byA i.e.|k(x, t)|


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wherek2(x, t) =b

a k(x, s)k(s, t)ds. Changing the variablettos in the fourth relation and usingthe expression in which we change variables. We have

3(x) = b

a

k(x, s)2(s)ds= b

a

k(x, s) b

a

k2(x, s)f(t)dtds=

ba

f(t)

ba

k(x, s)k(s, t)ds

dt=

ba

k3(x, t)f(t)dt

wherek3(x, t) =b

a k(x, t)k2(x, t)ds. Continuing the process and introducing the function

kn(x, t) = b

ak(x, s)kn1(s, t)ds

we obtain

n(x) = b

a kn(x, t)f(t)dt

The functionsk2(x, t), k3(x, t), . . . , kn(x, t)are referred to asiterated (repeated) kernels.

The kernelk(x, t)is the first kernel. Substituting the expression for n(x)into the approximateseries fory(x), we have

y(x) = f(x) +

ba

k(x, t)f(t)dt+ b

ak2(x, t)f(t)dt+ +

n1 b

akn(x, t)f(t)dt+

If the sequence

k1(x, t) +k2(x, t) +2k3(x, t) + +

n1kn(x, t) +

uniformly converges, then the sum in the square brackets in the expression for y(x) can bereplaced by the integral sum and we write

y(x) = f(x) + b

a(x, t,)f(t)dt

where

(x, t,) =k1(x, t) +k2(x, t) +2k3(x, t) + +

n1kn(x, t) +

This function

(x, t,)is called the Resolvent of the IE. When the Resolvent is known we canfind the solution of the IE for any function f(x)provided the parameteris sufficiently small inabsolute value.

Example

Evaluate, using the first three successive approximations of the solution of the IE below

y(x) +

10

xty(t)dt= x2

Solution

Suppose the solution has the form

y(x) =y3(x) =0(x) +1(x) +22(x) +

33(x)


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where

0(x) = f(x)

n(x) = 1

0

k(x, t)n1(t)dt we have

0(x) = x2

1(x) =

10

xt.t2dt=x

4

2(x) = 1

0xt.

t

4dt=

x

12

3(x) =

10

xt.t

12dt=

x

36

Since= 1, then the first three successive approximation is the approximate solution of theIE. Hence

y0(x) = 0(x) =x2

y1(x) = 0(x) 1(x) =x2

x

4

y2(x) = 0(x) 1(x) +2(x) =x2 x

1

4

1

4.3

y3(x) = 0(x) 1(x) +2(x) 3(x) =x2 x

1

4

1

4.3+

1

4.32

Example

Find the Resolvent, defining the radius of convergence of the sequence. Write out of the solution

for any arbitrary free term f

(x

)and also find the solution of the IE

y(x= f(x) + b

aexty(t)dt

Solution

Find the iterative kernels:

k1(x, t) = k(x, t) =ext

k2(x, t) = 1

0k(x, s)k1(s, t)ds=

ba

exsestds=ext

k3(x, t) = 1

0k(x, s)k2(s, t)ds=

b

aexsestds=ext

Hence

kn(x, t) =ext (n=1, 2, 3, . . .)

The Resolvent kernel is given by

(x, t,) =i

n f t yn=1n1kn(x, t) =exti

n f t yn=1n1

The series obtained is a geometric progression, which converges for ||


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The solution of the equation has the form

y(x) = f(x) +

1

10

extf(t)dt

In particular, for= 12

and f(x) =ex, we have

y(x) =ex +12

1 12

10

extetdt= 2e2

Exercises

1. Find the resolvent, radius of convergence of the series and the solution of the equation for

any arbitrary free term f(x)and also the solution for a given and f(x)of the followingequations

y(x) = 1

0 xty(t)dt+ f(x): =

1

2 ,f(x) =

5

6x

y(x) =

10

ty(t)dt+f(x): =1

2,f(x) =x

2. Solve the IE below by separating the kernel

y(x) =sinx +

0e(x+t)y(t)dt

wherealphais a real constant, and show that the solution is not valid when alpha=0andalpha= 1

2

3. Obtain the eigenvalues and eigen functions of the following equations:

y(x) =

11

(x t)y(t)dt

y(x) =

cos2 (x + t)y(t)dt


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2. Volterras IE

Successive Approximation for Volterras IE

Let us consider a Volterra equation of the second kind

y(x) = f(x) + x

ak(x,y)y(t)dt (3.1)

where f(x)is a continuous function in the interval [0, a]; k(x, t) is a continuous kernel for0 x a, 0 t x.

There are two possible forms of solution here. A situation where the solution is based on

iterating the unknown function y(t) and the other in which case the iteration is based on thekernel,k(x, t).

Iteration based on the Unknown

From the interval[0, a], we take a continuous function0(x)and replace it in the right hand sideof equation (3.1) instead ofy(t). we then obtain forn approximations

y(x) yn=0(x) +1(x) +22(x) + +

n1n1(x) +nn(x)

where

0= f(x) n= x

a

k(x, t)n1(t)dt

The sequence of functions yn(x), for n converges to the exact solution y(x) of theintegral equation (3.1). The functionyn(x)is an approximation of the solution of the equation.

Iteration based on the Kernel

The function could also be expressed by iterations of the kernel,k(x, t)where

k0(x, t) = k(x, t) and

kn(x, t) =

xa

k(s, t)kn1(s, t)ds (n=1, 2, 3, . . .)

The solution can then be considered as

yn(x) = f(x) + x

a

n

m=1

mkm+1(x, t)

f(t)dt


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10 Chapter 2. Volterras IE

We shall discuss the solution in terms of the a function called the Resolvent kernel, given

by

(x, t,) =

m=0

mkm+1(x, t)

The solution can then be written in terms of the resolvent kernel as

y(x) = f(x)

xa(x, t,)f(t)dt

Let us consider examples of both cases that have been discussed in this section

Example 3.1

Evaluate the first three approximations of the solution of Volterras IE second kind

y(x) + x

0(x t)y(t)dt= 1

Solution

In a general form we have

y(x) =1

x0

(x t)y(t)dt

f(x) =1,= 1, k(x, t) = (x t)For the first three approximations

y(x) y3(x) =0(x) +1(x) +21(x) +

33(x)

0 = f(x) =1

n = x

0k(x, t)n1(t)dt (n=1, 2, 3, . . .)

1 = x

0

(x t)0(t)dt= xt1

2

t2x

0

= x2 1

2x2 =

x2

2

2(x) =

x0

(x t)1(t)dt=

x0

(x t)t2

2dt

= xt3

2.3

t4

2.4

x

0

=x4

4!

3(x) =

x0

(x t)2(t)dt=

x0

(x t)t4

4dt

= xt5

4!.5

t4

6.4!x

0

=x6

6!Consequently the approximate solution for the given equation where = 1 is

y0(x) 1

y1(x) 1 x2

2

y2(x) 1 x2

2 +

x4

2

y3(x) 1 x2

2 +

x4

2

x6

6!

In the next example we shall use the kernel as an approximation. This helps us to first findthe resolvent kernel and then the eventual solution.


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11

Example 3.2

For Volterras IE of the second kind given below, find the resolvent kernel and the solution.

y(x) = x

t

exty(t)dt+ f(x)

Solution

We first find the iterative kernel

k0(x, t) = k(x, t) =ext

kn(x, t) = x

tk(x, s)kn1(s, t)ds

k1(x, t) = x

tk(x, s)k0(s, t)ds=

xt

exsestds

= (x t)ext

k2(x, t) = x

t k(x, s)k1(s, t)ds= x

t e

xs

(s t)est

ds

= 1

2(x t)2ext

k3(x, t) =

xt

k(x, s)k2(s, t)ds=

xt

exs1

2(s t)2estds

= 1

3!(x t)3ext

kn(x, t) = (x t)n

n! ext

The resolvent kernel is

(x, t,) =

m=1

mkm+1

=

m=1

m(x t)m

m! ext =e(xt)e(xt)

= e(+1)(xt)

Substituting this into the main equation, we have

y(x) = f(x) + x

t(x, t,)f(t)dt

= f(x) + x

t

e(+1)(xt)f(t)dt

The inconvenience of the method of successive approximation is the necessity to calculate the

quadrature. If this is not possible then one may use numerical methods.

Exercise

1. Solve completely

y(x) =5

6+

1

2

x0

xty(t)dt

2. For the above integral equation, use the method of successive approximation to solve

3. from 1. and 2. above, estimate the accuracy of the results


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3. Fredholms IE: Degenerate Kernels

Introduction

The main objective of this section is to find a solution to Fredholms IE of the second kind not

only for a sufficiently small value of the parameterbut also forfor which the solution exists.

Defn: Degenerate kernel

The kernel k(x, t) of an IE is called degenerate if it can be represented in the form of a finitesum of pairwise product of functions; one of which is a function of only x and the other being a

function of onlyt.

Mathematically,

k(x, t) = a1(x)b1(t) + a2(x)b2(t) + a3(x)b3(t) + + an(x)bn(t)

=n

i=1

ai(x)bi(t)

whereai(x), bi(t)are linearly independent functions.

Substituting this into Fredholms IE of the second kind,

y(x) = f(x) +

ba

k(x, t)y(t)dt

= f(x) +

b

a

ni=1

ai(x)bi(t)

y(t)dt

= f(x) + b

aa1(x)b1(t)y(t)dt+

ba

a2(x)b2(t)y(t)dt+ + b

aan(x)bn(t)y(t)dt

= f(x) +a1(x) b

ab1(t)y(t)dt+a2(x)

ba

b2(t)y(t)dt+ +an(x) b

abn(t)y(t)dt

= f(x) +n

i=1

ai(x) b

abi(t)y(t)dt

by letting

ci=

b

abi(t)y(t)dt (i=1, 2, 3, . . . , n)


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14 Chapter 3. Fredholms IE: Degenerate Kernels

whereci are constant coefficients, we can rewritey(x)as

y(x) = f(x) +n

i=1

ai(x)ci

Therefore if the coefficients ci are found then the problem is considered solved. This is

possible by substitutingy(x)into the function involving ci.Therefore we have

ci =

ba

bi

f(t) +

n

j=1

cjaj(t)

dt

ci = b

abi(t)f(t)dt+

ba

bi(t)n

j=1

cjaj(t)dt

ci n

j=1

cjgi j = fi

where

gi j =

ba

aj(t)bi(t)dt

fi = b

abi(t)f(t)dt

or in a more general representation as

n

i,j=1

(i j gi j) ci= fi

wherei j is Kroneckers symbol such that

i j=

1 i= j;0 i = j

This represents a system for iterating values ofi and j. We have

c1 = n

j=1

cjg1j

c2 = n

j=1

cjg2j

......

...

cn = n

j=1

cjgn j

(1 g11)c1 g12c2 g1ncn = f1

g21c1 (1 g22)c2 g2ncn = f2

g31c1 g31c2 g3ncn = f3...

. ..

...

..

.

..

.gn1c1 gn1c2 (1 gnn)cn = fn


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15

The matrix representation of the system being

(1 g11) g12 g1ng21 (1 g22) g2n

g31 g31 g3n...

. . . ...

...

gn1 gn1 (1 gnn)

c1c2

c3...

cn

=

f1f2

f3...

fn

The determinant of the system is given by

D() =

(1 g11) g12 g1ng21 (1 g22) g2ng31 g31 g3n

... . . .

......

gn1 gn1 (1 gnn)

LetAi j()be the algebraic adjoint of the corresponding element(or co-factors)i j gi j ofthe determinantD().

IfD() =0, then byCramers Rule,

ci=Dj()

D() =

ni,j=1Ai jfi

D()

where Dj() is the determinant of the system of equations. The solution of the system ofequation is therefore

y(x) = f(x) + b

a

ni=1

nj=1 ai(x)bi(t)Ai j()f(t)

D() dt

We defineD(x, t,) = ni=1nj=1 ai(x)bi(t)Ai j(). Then we can write the solution as

y(x) = f(x) + b

a

D(x, t,)

D() f(t)dt

The Resolvent kernel

(x, t,) = D(x, t,)D()

and finallyy(x)is expressed in terms of the resolvent kernel as

y(x) = f(x) + b

a(x, t,)f(t)dt

Example

Solve the IE

y(x) 2 1

0(1 + 3xt)y(t)dt= x2


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Solution

y(x) = 2 1

0(1 + 3xt)y(t)dt+x2

= 2

1

0y(t)dt+ 6x

1

0ty(t)dt+x2

= 2c1+ 6xc2+x2

wherec1=1

0 y(t)dtandc2=1

0 t y(t)dt

y(t) =2c1+ 6tc2+t2

Substituting this into c1, c2 we have

c1 = 1

0 2c1+ 6tc2+ t2

dt= 2c1

10

dt+ 6t

10

td t+

10

t2dt

c1 3c2 = 1

3 ()

Similarly,

c2 = 1

0t

2c1+ 6tc2+ t2

dt

= 2c1 1

0

tdt+ 6t 1

0

t2dt+ 1

0

t3dt

c1 2c2 = 1

4 ()

Combining()and()in a system, we have 1 31 2

c1c2

=

1

314

Therefore using Cramers Rule to solve the system,

c1=

13 314

2 1 31 2

= 1

12

c2=

1 131 14

1 31 2

= 1

12

The solution therefore is

y(x) =x2 12

x 16


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17

Example

Solve the IE

y(x) 1

0

(1 +x + t)y(t)dt= f(x)

Solution

y(x) =

10

(1 +x + t)y(t)dt+f(x)

= (1 +x) 1

0y(t)dt+

10

ty(t)dt+ f(x)

= (1 +x)c1+c2+f(x)

wherec1=

1

0 y(t)dtandc2=

1

0 t y(t)dt

We then substitute the functiony(x)intoc1, c2 we have

c1 =

10

((1 + t)c1+c2+f(t))dt

c1 = c1

10

(1 + t)dt+c2

10

dt+ 1

0f(t)dt

(1 3

2)c1 c2 =

10

f(t)dt ()

Similarly,

c2 = 1

0t((1 + t)c1+c2+f(t))dt

c2 = 5

6c1+

1

2c2+

10

t f(t)dt

5

6c1+ (1

1

2)c2 =

10

t f(t)dt ()

We put equations()and()together in a system as follows:

(1 3

2)

56 (1 1

2)

c1c2 =

10 f(t)dt

10 t f(t)dt The determinantD()is computed as

D() =

(1 32) 56 (1 1

2)

= (1

1

2)(1

3

2) +

5

62

= 1 2+ 1

122

Solving for the coefficients c1, c2 using Cramers Rule,

c1=D1()

D( c2=

D2()D(


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D1() =

1

0 f(t)dt

1

0 t f(t)dt (1 12)

= (1 1

2)

1

0f(t)dt+

1

0t f(t)dt

= 1

0

1

1

2+ t

f(t)dt

c1 = 1

0

1 1

2+ t

f(t)

1 2 1122

dt

D2() = (1 3

2)

1

0 f(t)dt

56 1

0t f(t)dt

= (1 3

2)

10

t f(t)dt+5

6

10

f(t)dt

=

10

t

3

2+

5

6

f(t)dt

c2 =

10

56+ t 3

2t

f(t)

1 2 1122

dt

The solution is arrived at by substituting the expressions for c1 and c2 into

y(x) =(1 +x)c1+c2+f(x)

Therefore we have

y(x) =(1 +x) 1

0

1 1

2+t

1 2 1

122

dt+ 1

0

56+ t 3

2t

1 2 1122

dt+f(x)

Finally we can simplify the solution as

y(x) =

10

1 1

2+t

+

56+ t 3

2t

f(t)

1 2 1122

dt+ f(x)

Example

Find the solution of the IE with degenerate kernel

y(x)

x cost+ t2 sinx + cosx sin t

y(t)dt= 2x

Solution

y(x) =

x cost+ t2 sinx + cosx sin t

y(t)dt+ 2x

= x

costy(t)dt+sinx

t2y(t)dt+cosx

sinty(t)dt+ 2x

= xc1+sinxc2+cosxc3+ 2x


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19

where

c1 =

cos ty(t)dt

c2 =

t2y(t)dt

c3 =

sin ty(t)dt

Substituting the functiony(x)intoc1, c2,and c3 we have

c1 =

tcost[tc1+sin tc2+cos tc3+ 2t]dt

c1 = c1

costdt+c2

costsin td t+c3

cos2 tdt+ 2

tcostdt

c1 c3 = 0 ()

Similarly,

c2 =

t2 [tc1+sin tc2+cos tc3+ 2t]dt

c2 = c1

t3dt+c2

t2 sintdt+c3

t2 cos td t+ 2

t3dt

c2+ 4c3 = 0 ()

Lastly,

c3 =

sin t[tc1+sin tc2+cos tc3+ 2t]dt

c3 = c1

tsin td t+c2

sin2 td t+c3

sin tcos td t+ 2

tsin td t

c3 2c1 c2 = 4 ( )


20/21


Aside(Integrals)

t3dt =

t4

4

= 4

4

()4

4 =0

sin2 tdt =

1

2(1 cos2t)dt=

1

2

t

1

2sin2t

= 1

2

+

1

2sin2

()

1

2sin (2)

=

cos2 tdt =

1

2(1 + cos2t)dt=

1

2

t+

1

2sin2t

= 1

2

+

1

2sin2

() +

1

2sin (2)

=

sintcos tdt = 1

2

sin (2t)dt= 1

4 cos (2t)|

= 14(cos (2) cos (2)) =0

sintdt = cos t|

= [cos cos ()] =0

tsintdt = tcos t|+

cos tdt

= [cos+cos ()] +sin t|

= 2+ sin sin() =2

t2 cos2 tdt = 0

t2 sin2 tdt = 4

Solving the three equation(), ()and ( )in a system, we have:

(1 0 0 1 4

2 1

c1c2

c3

=

00

4

We calculate the determinantD()is computed as

D() = 1 0 0 1 4

2 1

= 1 222 + 422

= 1 + 222

D1() =

0 0 0 1 4

4 1

= (4)= 42


21/21

21

D2() =

1 0 0 0 4

2 4 1

= 1(162) (0)

= 162

D3() =

1 0 0

0 1 0

2 4

= 1(4 0)

= 4

This means

c1= D1

D()=

42

1 + 222 c2=

D2

D()=

162

1 + 222 c3=

D3

D()=

4

1 + 222

By substituting the expressions for the constant coefficients into the equations,

y(x) = 4

1 + 22(4sinx + cosx) + 2x

Exercises

1. y(x) 4

2

0

sin2 t

y(t)dt= 2x

2. y(x) 1

1earcsinxy(t)dt= tanx

3. y(x)

4

4

(tant)y(t)dt= cotx

Integral Equation

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