Page 282 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 18: Charpit s Method for Solving Partial Differential Equations

Notes
1 2 2 a 2 1
2
2
2
(a + 1) z = t + [t (a 1)] log [t + (t (a + 1)}] + b
2 2
which is the required solution where t = ax + y.
Example 8: Solve by Charpit s method:
2
q = (z + px) .
Solution:

Applying Charpit s method,
dp dq dz
=
F F F F F F
p q q
x y y z p q

dx dy df
= .
F F 0
p q

We have
dp dq dx
2 (z px ) p 2(z px ) 2 (z px ) 2 (z px )
x
q
p
dq dx
or
q x
or qx = a

a 2
Putting this value of q in the given equation (z px )
x

1 a
or p z .
x x
Now dz = p dx + q dy
1 a a
= z dx dy
x x x
a
or (x dz + z dx) = dx + a dy
x
or zx = 2 (ax) + ay + b.

2
2
Example 9: Solve p + q 2px 2qy + 2xy = 0.
Solution:
Applying Charpit s method,
dp dq dz dx dy
=
F F F F F F F F
p q p q
x z y z p q p q

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