Page 333 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes
)
and (1 q dy (1 p )dx = 0 ...(4)
From (3), and
dz = p dx q dy , we have

dz = 0, or z = C ...(5)
1
and from (4), and
dz = p dx + q dy, we have
dx + dy + dz = 0,

or, x + y + z = C ...(6)
2
Now combining (3) with (1)

(q 1)dp (p 1)dq = 0 ...(7)
and combining (4) with (1),
q dp p dq = 0 ...(8)

i.e., dp dq = 0 [from (7) and (8)]
or p q = k 1 1 (C 1 ) 1 ( )
z
dx dy dz
=
1 1 1 ( )
z
z
z
or x = F 1 ( ) k 2 F 1 ( ) F 2 (C 2 )
= F 1 ( ) F 2 (x y z )
z

2
2
Example 5: Solve : q r 2pqs p t 0 and show that the integral represents a surface
generated by straight lines which are parallel to a fixed plane.
Solution: Putting
dp s dy dq s dx
r = , and t ,
dx dy

2
2
2
2
(q dpdy p dq dx ) = s (q dy 2 2pq dx dy p dx 2 )
The subsidiary equations are
2
2
q dp dy p dq dy = 0 ...(1)
q dy p dx = 0 ...(2)

Also dz = p dx q dy 0.

z = c.
From (1) and (2),
or q dp p dq = 0

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