Page 53 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes By putting these values in (i) we have
1
(1 2xh h 2 ) 1/2 = 1 xh h (x 2 1) cos 1 d
0
1/2
1 i 1
x
or 1 2 . 2 = { h x (x 2 1) cos 1 d
h h h 0
1 1
or P n ( ) = (t 1) d where t h x (x 2 1) cos
x
h h n 0
n 0
1 1 1
= 1 d
0 t t

1 1 1 1
= 1 2 ... n ... d
0 t t t t
1 1 1 1
= 2 3 ... n 1 d
0 t t t t

1
= d
0 t n 1
n 0
d
=
0 n 1 2 n 1
n 0 h x (x 1) cos

1
,
Equating the coefficient of n 1 we get
h
d
P n ( ) = ]
x
0 2 n 1
x (x 1) cos
1 d
x
P n ( ) =
x 0 2 n 1
x (x 1) cos
Deductions: Replacing n by (n 1) in above relation, we get

1 d
P (n 1) ( ) =
x
0 2 n
x (x 1) cos
1 x
= x (x 2 1) cos d
0
= P n ( )
x
x
P n ( ) = P n 1 ( )
x

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