Page 455 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes
x
u
x
or y ( ) y (0)x 9 y ( )du (x u ) = 3x 3 ...(2)
0
Putting x = /2 in (2) and using boundary condition we have
/2
u
y (0) 9 u y ( )du = 3 3
2 2
0
Solving for y (0), we have

/2
18 2
y (0) = u y ( )du 6 ...(3)
u
2
0
Substituting in equation (3) we have
/2 x
18x 2
u
x
y ( ) u y ( )du 6 x 9 (x u ) ( ) = 3x 3
y
u
2
0 0
x /2 x
18x 18x 3 2
y
u
or y ( ) u y ( )du u y ( ) 9 (x u ) ( ) = 3x 6 x ...(4)
u
u
x
2 2
0 x 0
or letting
F(x) = 3x (x + 2 ) ...(5)
2
2
18x
u 9(x u ) for u x
2
and K(x, u) = ...(6)
18x
u for u x
2
Equation (4) then becomes
/2
x
x
y ( ) K ( , ) ( ) = F(x) ...(7)
u
y
u
0
which is the required integral equation of the second kind.
Self Assessment
1. Convert the differential equation
x
y
y 4y sin3 with (0) y (1) 0
into integral equation.

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