Page 332 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

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

Notes
2
3
x ax
c
or p bq (y ax ) = 1 ( ) (y ax )
1
2 3
x 2 bx 3
p aq (y bx ) = 2 ( ) 2 (y bx ).
c
2
2 3
Solving,
1 yx 2 2 2 x 2
p = (a b ) (a b ) a 1 (y ax ) b 2 (y bx ) ,
a b 2 6
1 x 3
q = (a b ) 1 (y ax ) 2 (y bx )
b a 6

Putting these values in dz = p dx + q dy,

yx 2 x 3 a (y ax ) a (y bx ) x 3 (y ax ) (y bx )
dz = (a b ) 1 dx 2 1 2 dy
2 6 a b a b 6 a b a b

3
2
(a b )x 3 3x y dx x dy 1 1
)(
= dx [ ( y ax dy a dx )] [ 2 ( y bx dy b dx )]
)(
1
6 6 a b a b
(a b )x 3 yx 3
z = 1 (y ax ) 2 (y bx ).
24 6
Note: This question could be solved by the method of Ist chapter also.
Example 4: Solve by Monge’s method

q (1 q )r (p q 2pq )s p (1 p )t = 0.

Solution: Putting
dp s dy dq s dx
r = , t .
dx dy

2 dp sdy dq sdx
(q q ) (p q 2pq )s p (1 p ) = 0
dx dy

or [(q q 2 )dp dy (p p 2 )dq dx ]

= s [(q q 2 )dy 2 (q q 2pq )dx dy (p p 2 )dx 2 ]

The subsidiary equations are
(q q 2 )dp dy p (1 p )dq dx = 0 ...(1)

and [(q q 2 )dy 2 (p q 2pq )dxdy (p p 2 )dx 2 ] = 0 ...(2)
From (2), q dy + p dx = 0 ...(3)

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