Page 284 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 18: Charpit s Method for Solving Partial Differential Equations

Solution: Notes
Here f = pxy + pq + qy yz = 0 ... (1)
Charpit s auxiliary equations are

dp dq
= ...
py ( p y ) (px q ) qy
or dp = 0 or p = a ... (2)

From (1) and (2), we get
( y z ax )
p = a, q =
a y
Putting these values of p and q in dz = p dx + q dy, we get

( y z ax )
dz = a dx + dy
a y

dz a dx y dy a
or = 1 dy
z ax a y a y

Integrating, log (z ax) = y a log (a + y) + log b
y
a
or (z ax) (y + a) = be .
Example 11: Solve by Charpit s method:
px + qy = z(1 + pq) 1/2 .
Solution:

f = px + qy z (1 + pq) 1/2 = 0 ... (1)
Charpit s auxiliary equations are

dp dq
= ...
p p (1 pq ) 1/2 q q (1 pq ) 1/2
dp dq
or p = aq ... (2)
p q

Putting in (1), we get
2 1/2
q (ax + y) = z (1 + aq )
2
2
2
or q [(ax + y ) az ] = z 2
z az
q = 2 2 1/2 and p = aq = 2 2 1/2
[(ax y ) az )] [(ax y ) az ]
putting these values of p and q in dz = p dx + q dy,

( z a dx dy ) dz a dx dy
dz = or
{(ax y ) 2 az 2 } z {(ax y ) 2 az 2 }

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