Page 318 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

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

Notes
1 2 2 2 2
x
= tan x tan(c x ) tan tan (c x ) {sec x sec (c x )}dx
2(D D )
1 2 2
x
= tan x tan(c x ) tan tan (c x ) tanx tan(c x )
2(D D )
1 2 2
x
= tan x tan y tan tan y tanx tan y [By putting back y = c x]
2(D D )
1 2 2
y
x
= 2(D D ) [tan sec x tan sec y ]
1 2 2
= [tan(k x )sec x tan sec (k x )]dx where k + x = y
x
2
1 d
x
= {tan tan(k x )} dx
2 dx
1 1
x
x
= tan tan(k x ) tan tan y [putting k + x = y]
2 2
Hence the complete solution is
1
x
z = (y x ) 2 y x tan tan y
2
Example 4: Find the particular integral with the help of general method for

(D 2 2DD 15D 2 )z 12xy

Solution: We have
1
P.I. = 12xy
(D 2 2DD 15D 2 )

1
= 12xy
(D 3D )(D 5D )

12
x
= ( x c 5 )dx , where y = c 5x
(D 3D )
12 cx 2 5x 3
=
D 3D 2 3

2 2 3
= (3cx 10x )
D 3D
2 2
x
= x (3y 15x 10 ), (putting back c = y + 5x)
D 3D

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