Page 308 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 20: Higher Order Equations with Constant Coefficients and Monge’s Method

4. Solve Notes
2 2
z 6 z 9z 0
x 2 x y

20.3 The Particular Integral (P.I.)

We now return to the equation (3) i.e.

F ( ,D )z = f ( , ) ...(1)
y
x
D
Now the most general solution of equation (1) can be written as
z = complementary function + Particular function
or z = C.F + P.I ...(2)
In the above we have found C.F. for the homogeneous equation and now in the following find
the P.I. We can write

1
y
The particular integral = f ( , ) ...(12)
x
D
F ( ,D )
Here we treat the symbolic function of D and D as we do D alone. We can factor. ( ,F D D ,
)
1
resolve into partial fractions on expanding in power series.
D
F ( ,D )
(a) On Expansion

Example 1: Solve

(D 2 4DD 4D 2 )z = 0

The complementary function is given by
(D 2 4DD 4D 2 )z = 0

x
x
C.F. = F 1 (y 2 ) x F 2 (y 2 )
The particular integral is

1 2
P.I. = 2 2 (x xy )
D 4DD D (4)
1
or P.I. = (D 2 4DD 4D ) (x 2 xy )

1
1 4D D 2
= 1 4 (x 2 xy )
D 2 D D 2
1 4D 4D 2 16D 2 2
= 2 1 2 2 ... (x xy )
D D D D

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