Page 309 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes 1 4
= 2 x 2 xy ( ) 0
x
D D
3
x 4 x y 4
x
= 3 ( )
12 6 D
3
3
x 4 x y x 4 x 4 x y
=
12 6 24 8 6
Thus the complete solution is
3
x 4 x y
x
x
z = F 1 (y 2 ) x F 2 (y 2 )
8 6
Example 2: Solve

2
(D 2 a D 2 )z = x 2
Solution: The complementary function is given by the equation

2
(D 2 a D )z = 0
The auxiliary equation is

m 2 a 2 = 0
with roots m = a and m = a.

So C.F. = F 1 (y ax ) F 2 (y ax )
The particular integral is given by

1 2
P.I. = (x )
2
(D 2 a D 2 )
2
1 a D 2 1 2
= 2 1 2 (x )
D D
2
1 a D 2 1 x 4
= 1 ... x 2 (x 2 )
D 2 D D 2 12
So the complete solution is

x 4
z = F 1 (y ax ) F 2 (y ax ) .
12
Self Assessment

2 2 2
z z z
5. Solve 2 6 2 xy
x x y y
2 2 2
z z z
6. Solve 2 3 2 2 x y
x x y y

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