Page 340 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

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

Notes
2
2
Solution: Here R = z (1 q ), S 2pqz T (1 p )z
,
U = z 2 ,V (1 p 2 q 2 ).
The equation in is
(RT UV ) 2 US U 2 = 0

3
p q
or z 2 2 2 2 2 z pq z 4 = 0
2
or p q 2 2 2z pq z 2 = 0

or = z /pq . (roots are equal).
The system of intermediate integrals is given by
U dy T dx U dp = 0

U dx Rdy U dq = 0.

i.e., by pq dy (1 p 2 )dx zdp = 0

pqdx (1 q 2 )dy zdq = 0.

Also dz = p dx q dy .

We write (1) as
dx ( p pdx qdy ) zdp = 0,

With the help of (3), it reduces to
dx p dz z dp = 0

or x + pz = .
Similarly from (2) and (3), y zq .

Putting the values of p and q in dz p dx q dy ,

x y
dz = dx dy
z z
or - z dz = ( x )( dx ) ( y )( dy )

z 2 ( ) x 2 ( ) y 2
or = k
2 2 2

or z 2 (x ) 2 (y ) 2 = 2
Where , , are constants.

Example 3: Solve: (1 q 2 )r 2pqs (1 p 2 )t

(1 p 2 q 2 ) 1/2 (rt s 2 ) = (1 p 2 q 2 3/2 .
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