Page 316 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

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

Notes
1
x
x
x
= [(y 3 )sinx 3 sinx 3cos ]
(D 2D )
[putting back c 3x y ]
1
x
y
= [ sinx 3cos ]
(D 2D )
x
x
= {(k 2 )sinx 3cos }dx [ 2x k ] y
x
= k cosx 2( x cos x sin ) 3sin x

x
x
x
= (y 2 )cosx 2 cos x sin x [ y k 2 ]
x
= y cosx sin .
Hence the complete solution is
z = C . . P . . 1 (y 2 ) 2 (y 3 ) y cos x sin .
x
x
I
x
F
4x y
2
2
Example 2: Solve: (D 4D )z 2 2
y x
x
x
Solution: The C.F. = 1 (y 2 ) (y 2 )
1 4x y
Now, P.I. =
(D 2D )(D 2D ) y 2 x 2
1 4x c 2x

= 2 2 dx ( c 2x ) y
x
D 2D (c 2 ) x
1 2x c c c 2
= 2 2 2 dx
x
D 2D (c 2 ) x x
1 1 2c c 2
= 2 2 dx
D 2D c 2x (c 2 ) x x
x
1 c c
x
= log(c 2 ) 2log x
D 2D c 2x x
1 y 2x y 2x
= log y 2log x [putting c = y + 2x]
D 2D y x

k 4x k 4x
x
= log(k 2 ) 2log x dx where y = k + 2x
k 2x x
2x k k k
= log(k 2 ) 1 4 2log x dx
x
k 2x x

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