Page 420 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 24: Integral Equations and Algebraic System of Linear Equations

Notes
for (x x 0 ) . h Substituting this estimate for ( )y t z ( ) once more on the right side of (5) we
t
have
3
y ( ) z ( ) KN x x 0 / 3
x
x
for x x 0 . h Repeating this substitution we obtain
m
x
x
y ( ) z ( ) NK x x 0 / !, m 1,2,...
m
for x x 0 . h The right side of the above inequality tends to zero as m . This means that
N = sup y ( ) z ( )
x
x
x 0 x h
is equal to zero. So the solution of y(x) by the integral equal is unique also.
24.2 Conversion of a Differential Equation of Second Order to an
Integral Equation

Example: Convert the differential equation

2
d y dy 2
2 8y = 5x 3x ...(1)
dx dx
with the initial conditions

dy
x = 0, (y x 0 ) 2, 3. ...(2)
dx x 0
Solution 1: Let
2
d y
x
y = 2 G ( ) ...(3)
dx
Integrating (3) once yields the result

x
x
y ( ) = G ( ) dt C 1
t
0
For x = 0, this gives
y (0) = 0 + C = 3
1
therefore
x
y ( ) = G ( )dt 3 ...(4)
x
t
0
Again integrating (4),

x t x
y(x) = G ( ) dt dt 3 dx C 2
t
0 0 0

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