Page 421 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes Integrating the first term on the right by parts we have
x
t x
t
t
y(x) = t G ( )dt tG ( )dt 3x C 2
0 0 0
x x
t
t
= x G ( )dt t G ( )dt 3x c 2
0 0
x
or y(x) = (x t ) ( )dt 3x c
t
G
2
0
Subjecting this to the condition
y(x) = 2 at x = 0
we get 2 = 0 + 0 +c or c = 2
2 2
x
t
so y(x) = (x t ) ( )dt 3t 2 ...(5)
G
0
Writing (1) with the help of (3), (4) and (5), we have
x x
G ( ) 2 G ( )dt 6 8 (x t ) ( )dt 24t 16 = 5x 2 3x
x
G
t
t
0 0
x
t
t
or G ( ) (2 8x 8 ) ( )dt 5x 2 21t 22 = 0 ...(6)
G
x
0
2
d y
Where G(x) = 2 ...(7)
dx
Solution 2: We follow an other method. In this method we integrate equation (1) from 0 to x,
x
x x 5 3 3 2
t
t
y
y ( ) 2 ( ) 8 y ( )dt = x x
t
0 0 3 2
a
x
5 3
y
x
y
or y ( ) y (0) 2 ( ) 2 (0) 8 y ( )dt = x 3 x 2
t
x
3 2
0
but y (0) = 3, y(0) = 2
y (0) 2 (0) = 1
y
x
5 3 3 2
y ( ) 2 ( ) 8 y ( )dt = x x 1
x
y
x
t
3 2
0
Again integrating, we get
x x 3
x 5 4 x
y ( ) 2 y ( )dt 8 (x t ) ( )dt = x x
t
t
t
y
0 12 2
0 0
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