Page 278 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 18: Charpit s Method for Solving Partial Differential Equations

Any of the integrals of (7) will satisfy (6). The simplest relation involving p or q or both should Notes
be taken and that will be the required relation.

18.2 Illustrative Examples

Example 1: Solve by Charpit s method z = pq.
Solution:

Applying Charpit s method,
dp dp dz dx dy df
=
p . 1 q ( p )( q ) ( q )( p ) q p 0
From first two terms,

p
= c.
q

2
z = cq or q = (z/c) and p = (cz).
Now dz = p dx + q dy
= (cz) dx + (z/c) dy
z 1/2 dz = c dx + (1/ c) dy, on integration, we have

2z 1/2 cx + (y/ c) + b

2
2
Example 2: Solve by Charpit s method (p + q ) y = qz.
Solution:

dp dq dz dx dy df
0 p q ) (p 2 q 2 ) q q ) p (2py ) q (2qy z ) 2py 2py z 0
(
(
From first two terms,

dp dq
qp p 2

2
2
or p dp = q dq i.e. p + q = c
2 2
2
 q cy/z and p = (c c y /z )
dz = pd x + q dy
2
2 2
= (c c y /z ) dx + cy/z dy
2 2 1/2
2
or z dz = (cz c y ) dx + cy dy
2(z dz cy dy )
or = 2 c . dx,
(z 2 cy 2 )
2
2 1/2
(z cy ) = c .x b

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