Page 330 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

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

Notes
1
and q = [ (y ax ) (y ax )]
1
2a
Substituting these values in dz pdx q dy , we have
1 1
dz = [ (y ax ) 2 (y ax dx [ (y ax ) 2 (y dx )]dy
)]
1
1
2 2a
1 dy a dx
or dz = (dy a dx ) 1 (y ax ) 2 (y ax ),
2a 2a
or z = f 1 (y ax ) f 2 (y ax ).

Example 2: Solve by Monge’s method:
2
2
)
)
(b cq r 2(b cq )(a cp )s (a cp t = 0.
Solution. Putting
dp s dy dq s dx
r = , t ,
dx dy
2 dp s dy 2 dq sdx
(b cq ) 2(b cq )(a cp )s (a cp ) = 0.
dx dy

The subsidiary equations are,
2
2
(b cq ) dy 2 2(b cq )(a cp )dx dy (a cp dx 2 = 0, ...(1)
)
2
2
)
(b cq dp dy (a cp dx = 0, ...(2)
)
From (1),
(p cq )dy (a cp )dx = 0 ...(3)
Combining it with (2),

(b cq )dp (a cp )dq = 0

dp dq
From which =
a cp b cq

and therefore, (a cp ) = A (b cq ). ...(4)

Also from (3) and dz pdx q dy , we get
a dx b dy c dz = 0

or ax by az = B. ...(5)

From (4) and (5),
a + cp = (b cq ) (ax by cz )

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