Page 311 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes 2 2
So (D 2DD D )z = sin(2x 3 )
y
The auxiliary equation is

m 2 2m 1 = 0
having roots m = 1, 1, so that
C.F. = F 1 (y x ) x F 2 (y x )

1
y
and P.I. = 2 sin(2x 3 )
(D D )
Putting 2x 3y , u so we have

1
P.I. = sin u du du (integrating twice)
,
(2 3) 2
= 1 ( cos )u du

y
= sin u sin(2x 3 )
Thus the solution is

z = C.F. + P.I.
= F (y x ) xF (y x ) sin(2x 3 )
y
1 2
Example 2: Solve

(D 2 D 2 )z = 30(2x y )

The auxiliary equation is

m 2 1 = 0
so, m = +1, 1
and C.F. = F 1 (y x ) F 2 (y x )

1
P.I. = 30(2x y )
(D 2 D 2 )
Let u 2x y ,

1
P.I. = (30) (u du )du
(4 1)

1 u 2
= (30) du
3 2

u 3 5 3
= 10 (2x y )
6 6

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