Page 51 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes 1
Example 7: Prove that (1 2xz z 2 ) 2 is a solution of the equation

2 (z )
z (1 x 2 ) = 0
z 2 x x
Where
= (1 2xz z 2 ) 1/2
Solution: Let

n
= (1 2xz z 2 ) 1/2 z P
n
n 0
or z = z n 1 P
n
n 0
2
n
z (z ) = (n 1)nz P n .
z 2
n 0
n
Also = z P n
x
n 0
n
(1 x 2 ) = (1 x 2 ) z P n
x x x
n 0
n
n
= (1 x 2 ) z P n 2x z p n
n 0 n 0
Substituting this in the L.H.S. of the given equation, we get
2 (z )
n
. z 2 (1 x 2 ) = [(n 1)nz P (1 x 2 )z P 2xz P ]
n
n
z x x n n n
n 0
P
= z n [(1 x 2 )P n 2xP n ( n n 1) ]
n
n 0
P
= 0 since n is a solution of Legendre’s equation.
Self Assessment
2. Show that

1 z 2 x
x
P
z
n
(1 2xz z 2 3/2 = (2n 1) ( ) .
)
n 0
Laplace’s First Integral for P (x): when n is a positive integer. Show that
n
1 2 x
P n ( ) = x (x 1) cos d .
x
0
Proof: From integral calculus, we have
d
= , where a 2 b 2 ...(i)
0 a b cos 2 2
(a b )
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