Page 290 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 18: Charpit s Method for Solving Partial Differential Equations

On rearranging we have Notes

p 2 (1 x 2 ) q 2
x 2 y = a (say) ... (15)

ax
2
Then q = a y and p =
(1 x 2 1/2
)
Thus dz = pdx + qdy,
On integration gives

ax dx 2 y 2
z = a . b
)
(1 x 2 1/2 2
a 2 2
2 1/2
z = a (1 x ) y b ... (16)
2
is the complete integral.
(d) Clairaut s Equations

A first order partial differential equation of the form
z = px + qy + f (p, q) ... (17)
is of Clairaut type of the equation. Here

F = px + qy + f (p, q) z = 0 ... (18)
So from the corresponding Charpit s equations, we have
dp dq dz dx dy
, ... (19)
p p q q ( p x f p ) q (y f q ) x f p y f q

We have
p = a (say a constant)
q = b (a constant).

So from (17)
z = ax + by + f (a, b) ... (20)
is the complete solution of (17).

Example 5: Solve:
2
2
2
2
pqz = p (xq + p ) + q (yp + q ) ... (21)
Solution:
From (21)
p 3 q 3
z = px + qy +
q p

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