Page 322 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

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

Notes
1 x y
P.I. = [e ] x
(D D 1)(D D 2)
1 2x y 1
= e x
(D D 1)(D D 2) (D D 1)(D D 2)

1 2x y
Now, e
(D D 1)(D D 2)
1 y
2x
= e e
(D 2 D 1)(D 2 D 2)
1 y
2x
= e e
(D D 1)(D D )
1 y
2x
= e e
[0 ( 1) 1][0 ( 1)]
2 1 y 1 2x y
x
= e . e e
2 2
1
Also, x
(D D 1)(D D 2)
1
1 1
= [1 (D D )] 1 1 (D D ) x
2 2
1 1
= [1 D D ...] 1 (D D ) ... x
2 2

1 3 3
= 1 D D x
2 2 2

1 3 3 1 3
= x 0 x
2 2 2 2 4
The solution is

1 2x y 1 3
2x
x
z = e (y ) x e 2 (y ) x e x .
2 2 4
2 z 2 z z z
Example 3: Solve: 2 2 3 2z e x y x 2 . y
x y x y
2
Solution: [(D D )(D D ) 2(D D ) (D D ) 2]z e x y x y
2
or [(D D 2)(D D 1)]z e x y x y
The complementary function is
z = e 2x (y ) x e x (y ) x

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