Page 344 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

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

where Notes
t
f ( )
( ) = dt . ...(6)
t
1
2a
F ( )
t
and 2 ( ) = dt . ...(7)
t
2a
Hence the primitive is
z qy = 1 (q ax ) 2 (q ax )
y = 1 (q ax ) 2 (q ax ) [from (5), (6) and (7)].

Example 5: Solve:

rq (p x )s yt y (rt s 2 ) q = 0

),
Solution: Here R , q S (p x T , y U , y V . q
The equation in is

2 2
y
[qy qy ] . (p x ) y = 0
or = , or y /(p x ).

The intermediate integrals are given by

y 2 y
y dy dx dp = 0 ...(a)
p x p x
y
dx q dy y dq = 0

2
qy y dq
y dx dy = 0 ...(b)
p x p x
y
dy y dx y dp = 0

From (a)
y
[(p x )/ ] = ...(1)
qy = F( ) ...(2)
or one of the integrals is
y
qy = F [(p x )/ ].
From second equation of (b),
p x
p + x = , [from (1)]...(2 )
y y

p = x. ...(3)

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