Page 272 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 17: Lagrange’s Methods for Solving Partial Differential Equations

Notes
z a z 2
, (1 a ).
r r
z = a log r + a quantity independent of r
2
and z = (1 a ) + a quantity independent of .
General solution is

2
z = a log r + (1 a ) + c
y
2
1
2
= a log (x + y ) + (1 a 2 ) tan + c.
x
2
2
Example 4: Solve: (x + y) (p + q) + (x y) (p q) = 1.
Solution:
Put (x + y) = X, (x y) = Y,

z z X z Y z z
p . . .
x X x Y x X Y
z z X z Y z z
q . . ( 1).
y X y Y y X Y

On substitution the given equation becomes

2 2
z z 1
X Y
X Y 4
2 2
z 1 z
or X Y ,
X 4 Y
which is of the form f (x, p) = f (q, y).
1 2
2 2
z 1 z
Putting X = a and Y = a, we get
X 4 Y
z/ X = (a/X)

1
and ( z/ Y) = a /X .
4
z = 2 (aX) + a quantity independent of x

1
and z = 2 a Y + a quantity independent of y.
4

Complete integral is

1
z = 2 (aX) + 2 a Y + b
4

LOVELY PROFESSIONAL UNIVERSITY 265