Page 154 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 8: Total Differential Equations, Simultaneous Equations

Let the solution of (1) is Notes
Adu B dv ...(10)

then Adu Bdv = 0 ...(11)

is identical to (1) i.e.
A (dx dy dz ) B (z y )dx (x z )dy (y x )dz = 0 ...(12)

A ( B z y dx A ( B x z dy A ( B y x dz = 0 ...(12 )
)
)
)
Comparing (12 ) with (1) we have
A ( B y z ) y 2 yz z 2

A ( B x z ) z 2 zx x 2 ...(13)

A ( B x y ) x 2 xy y 2

From (13) we have B x y z u ...(14)

And A = (xy yz xz ) v ...(15)
Hence
Au Bv = 0 ...(16)

becomes vdu udv = 0

du dv
or = 0
u v
on integrating

u
log = log k
v
u
or = k ...(17)
v
From (6) and (9) we have

x y z
= k ...(18)
xy yz zx
which is the solution of equation (1).

Example 2: Solve

(2xz yz )dx (2yz zx )dy (x 2 xy y 2 )dz = 0 ...(1)
Solution: By the method of forming auxiliary equations

,
Here P 2xz yz Q 2yz zx R x 2 xy y 2
,

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