Page 449 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes So we have converted a differential equation (1) into the Volterra integral equation of the
second kind with Kernel given by (8) and the function given by (7). The unknown function being
given by (3).

Example: Convert

y
x
2y ( ) 3 ( ) 2 ( ) 4e x 2cos x ...(1)
y
x
x
with y(0) = 4, y (0) = 1, into integral equation
Let us put
y (x) = G(x) ...(2)
Integrating (1) with respect to x, we have
x
x
u
y ( ) G ( )du
x
0 0
dy x
u
or y (0) G ( )du
dx 0
dy x
u
1 G ( )du ...(3)
dx 0
Integrating with respect to x again we have
x u
x 1
x
u
y ( ) x du G ( )du
0 1
0 0
x
u
G
x
or y ( ) y (0) x (x u ) ( )du
0
x
G
u
y ( ) 4 x (x u ) ( )du ...(4)
x
0
Substituting from equations (2), (3) and (4) into (1) we have
x x
G
2 ( ) 3 1 G ( )du 2 4 x (x u ) ( )du 4e x 2 cos x
u
u
G
x
0 0
Rearranging we have
x
2 ( ) G ( )du 3 2(x ) u 4e x 2cos x 3 2 (4 ) x ...(5)
x
G
u
0
x
u
G
x
u
x
G
2 ( ) K ( , ) ( )du F ( ) ...(6)
x
0
where
K ( , ) 3 2(x ) u
x
u
...(7)
x
x
F ( ) 4e 2cos x 5 2x
So we get Volterra integral equation of the second kind.
Self Assessment
1. Convert the linear differential equation
3
d y
x
6 ( ) 0 with y (0) 4, y (0) 3, y (0) 2
y
dx 3
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