Page 314 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

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

So the complete solution is Notes

x 2
z = F (y ax ) x F (y ax ) ( f y ax )
1 2
2
Example 2: Solve

(4D 2 4DD D 2 )z = e x 2y x 3

Solution: The auxiliary equation is

4m 2 4m 1 = 0
m = 1/2, 1/2

C.F. = F 1 (2y x ) x F 1 (2y x )

1 x 2y 3
P.I. = 2 e x
(2D D )

1 1
= e x 2y x 3
(2D D ) 2 (2D D ) 2

x 2 x 2y 1 D 2 3
= e 1 x
2.4 4D 2 2D

x 2 x 2y 1 D
= e 2 1 ... x 3
8 4D D

x 2 x 2y 1 x 5 x x 2y x 5
So P.I. = e . e
8 4 4.5 8 80
Thus the solution is

x 2 x 2y x 5
z = F 1 (2y x ) x F 1 (2y x ) e
8 80
Self Assessment

9. Solve

(D D ) 2 x (x y )
10. Solve
2
(D 3 4D D 4DD 2 )z cos(y 2 )
x
20.5 General Method for Finding Particular Integral (P.I.)

Consider the equation

(D mD )z = f ( , )
x
y

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