Page 83 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes
On the infinite interval (0, ), we take as boundary conditions the following:
x
y ( ) remains finite as x 0,
y ( ) tends to infinity as 0(x ) as x .
x
The above equation when expanded is equal to
x
x
x e y e x (x 1)y e y = 0
or
x y (1 x )y y = 0 ...(i)

dy
Here y = .
dx
Equation (i) has only one finite regular singular point x 0 whereas x is irregular singular
point. So we can apply Frobenius method to express the solution of (i) as a power series:

y = a x k r ...(ii)
r
n 0
dy
= a (k r )x k r 1
dx r
r 0
2
d y
and 2 = a r (k r )(k r 1)x k r 2
dx
r 0
Substituting in (i), we get

a r (k r )(k r 1)x k r 1 (1 x )(k r )x k r 1 x k r = 0
r 0
or

2
a r (k r x k r 1 (k r )x k r = 0 ...(iii)
)
r 0
Now (iii) being an identity, we can equate the coefficients of various powers of x to zero.

Equating to zero the coefficient of lowest power of x, i.e., of x k 1 , we have

a k 2 = 0
0
Now, a 0 0, as it is coefficient of the first term with which the series is started.

k = 0.

Equating to zero the coefficient of general term, i.e., of x k r , we have

a (k r 1) 2 a (k r ) = 0
r 1 r

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