Page 254 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

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

So dx + m dy + n dz = 0 ... (3) Notes
On integrating (3) we have
x + my + nz = a = u (say) ... (4)
Again from (2)

xdx ydy zdz

( x mz ny ) ( y nx  ) z ( z y mx )

xdx ydy zdz xdx ydy zdz
or mxz nxy nxy  yz  zy mxz O

So xdx + ydy + zdz = 0
2
2
2
or x + y + z = b = (say) ... (5)
Hence the integral of (1) is
(u, ) = 0 ... (6)

Example 2: Solve

p q 1
x 2 y 2 zx

Solution:
The subsidiary equations are

dx dy dz
1 x 2 1 y 2 1 zx

2
2
or x dx = y dy = zxdz
From the first two equations we have on integration
3
3
x = y + a
3
3
or x y = a (say u)
From the first and third equations
2
x dx = xzdz
or xdx = zdz
On integrating it
2
2
x = z + b
2
2
or x z = b = (say b = )
So the solution of the above equation is
(u, ) = 0
3
3
2
2
(x y , x z ) = 0

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