Page 310 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

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

20.4 Shorter Method for Finding Particular Integral Notes

When dealing with the equation

x
F ( ,D )z = f ( , )
D
y
We consider a special function of the form
x
f ( , ) = (ax by ),
y
then a shorter method may be used. Now
D (ax by ) = a (ax by D (ax by ) b (ax by )
);
So D r (ax by ) = a r r (ax bx )

D r (ax by ) = b r r (ax bx )
p
p
and D D q (ax by ) = a b q p q (ax by )
Here n is the nth derivative of with respect to ‘ax + by’ as a whole and n is the degree of
D
F ( ,D ).
Hence we will have

D
b
a
F ( ,D ) (ax by ) = F ( , ) n (ax by ) ...(13)
when n is the nth derivative of with respect to ‘ax + by’ as a whole and n is the degree of
F ( ,D ).
D
1
Operating by on both sides of (13) and dividing by ( , ),F a b we get
F ( ,D )
D
1 n 1
D
F ( ,D ) (ax by ) = F ( , ) (ax by ) ...(14)
a
b
provided
b
F ( , )  0.
a
1 1
u
Therefore 1 (ax b ) = 1 ( )du ...du
a
F ( ,D ) F ( , )
b
D
1
= nth integral of where u = ax + by ...(15)
F ( , ) 1
b
a
Example 1: Solve
y
(r 2s t ) = sin(2x 3 )
Solution:

2 z 2 z 2
Here r = 2 ,s ,t 2
x x y y

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