Page 342 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

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

From (1) and (2), Notes
(x ) p
= . ...(3)
(y ) q
Substituting in (1) the value of p as found from (3),

y
q =
2
[1 {(x ) 2 (y ) }]
Similarly from (3) and (2),

x
p =
2
[1 {(x ) 2 (y ) }]
Now, dz = pdx qdy

(x )dx (y )dy
or dz = 2 2
[1 {(x ) . (y ) ]

Integrating,
2
(z ) = [1 {(x ) 2 (y ) }] 1/2

2
or (z ) 2 = 1 [(x ) 2 (y ) ]
or (x ) 2 (y ) 2 (z y ) 2 = 1.

Example 4: Solve s 2 rt a 2

2
or rt s 2 = a .
Solution: Here R 0, S 0, T 0, U 1, V a 2 .

The equation in is
2 2
( a ) . 0 1 = 0
or = 1/a.

The two intermediate integrals are given by
1
dy dp = 0, ...(a)
a

1
dx dq = 0.
a
1
dy dp = 0, ...(b)
a
1
dx dq = 0.
a

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