Page 315 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes
z z
y
x
i.e m = f ( , )
x y
x
y
or p mq = f ( , ) ...(1)
z z
where p = andq .
x y
So Lagrange’s auxiliary equations (A.E.) are

dx dy dz
=
x
y
1 m f ( , )
From the first two fractions, we have
y = mx c ...(2)
From the first and last fractions
dz = f ( , )dx f ( ,c mx )dx
y
x
x
x
z = f ( ,c mx )dx
and after integration (c mx) is replaced by y because the P.I. does not contain any arbitrary
constant.

Now, the particular integral of

1 1 . 1 ... 1 f ( , )
y
x
f ( , ) =
y
x
f ( ,D ) D m D (D m D ) D m D
D
2
1
n
can be determined by the repeated application of the method given above.
Illustrative Examples
Example 1: Solve: r s 6t y cos x
Solution: The given equation can be written as
(D 2 DD 6D 2 )z = y cosx

A.E. is m 2 m 6 0, i.e., m 2, 3

x
x
C.F. = 1 (y 2 ) 2 (y 3 )
1
Now, P.I. = y cosx
(D 2D )(D 3D )

1

x
= . (c 3 )cos xdx [ y c 3 ]
x
(D 2D )
1
x
c
x
= [ sinx 3 sin x 3cos ]
(D 2D )
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