Page 288 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 18: Charpit s Method for Solving Partial Differential Equations

Notes
Example 2: Solve:
p + q = pq ... (1)

Solution:
p = a (constant)
so from (1)
a + q = aq

a
or q =
a 1
a
Thus dz = a dx + dy
a 1
a
given z = ax + y + b
a 1
which is the general solution.
(b) Equations not involving independent variables consider the partial equation of the
following type
f (z, p, q) = 0 ... (1)
which does not involve independent variables x, y.
From the subsidiary equations:

dp dq dz dx dy df
= ... (2)
f pf f q f pf qf f f 0
x z y z p q p q
Here the symbols used are

f f f f f
f x , f p , f z , f q , f y ... (3)
x p z q y
So from the first two fractions of (2) we have

dp dq
pf qf
z z
Integrating, we have
p = aq ... (4)
From equations (1) and (4) we can find p and q and the complete integral follows from the
relation.
dz = pdx + q dy ... (5)

Example 3: Find the complete integral of the equation
2 2
p z + q 2 = 1 ... (6)
As (6) does not involve x, y. So from the above method

q = pa ... (7)
1

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