Page 426 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 24: Integral Equations and Algebraic System of Linear Equations

Notes
x x
f
u
or y ( ) y (0) y (0)x (x u ) ( )du (x u ) ( )du = 0
y
x
u
0 0
x x
u
f
or y ( ) y (0)x (x u ) ( )du (x u ) ( )du = 0 ...(3)
u
x
y
0 0
To find the value of y (0), put x = 1 in equation (3), we get
1 1
u
u
0 y (0).1 (1 u ) ( )du (1 u ) ( )du = 0
f
y
0 0
so y (0) is given by
1 1
y (0) = (1 u ) ( )du (1 u ) ( )du ...(4)
u
y
f
u
0 0
Substituting this value of y (0) in equation (3) and rearranging terms we get
1 x 1 x
u
y
u
f
u
y(x) = x (1 u ) ( )du (x u ) ( )du (1 u ) ( )du (x u ) ( )du
u
f
y
0 0 0 0
x x 1 x
u
u
y
y
or y(x) = x (1 u ) ( )du (x u ) ( )du x (1 u ) ( )du (x u ) ( )du
u
y
f
u
0 0 x 0
x 1
f
u
u
f
x (1 u ) ( )du (1 u ) ( )xdu 0
0 x
Simplifying the above equation we have
x 1 x 1
u
u
f
y
y(x) = u (1 x ) ( )du x (1 u ) ( )du ( u x 1) ( )du x (1 u ) ( )du 0 ...(5)
y
f
u
u
0 x 0 x
Defining
( u x 1) u x
K(u, x) =
( x u 1) u x
We write equation (4) as
1 1
y(x) = K ( , ) ( )du K ( , ) ( )dx
x
u
y
x
u
u
u
f
0 0
Knowing K(u, x) and f(u), we know the second integral on the right hand side. Let us put
1
x
u
K ( , ) ( )du = (x) ...(6)
u
f
0
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