Page 147 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes 2 2 2 2 2 2
= 2yx (x y )[ x y yz y x xz 2x 2y yz xz ]
= 2xy (x y )[0] 0 ...(3)

So integrability condition is satisfied.
2
2
Now dividing by x y eq. (1) we have

1 y z 1 z x 1 1
dx dy dz = 0
y x 2 x 2 y y 2 y 2 x y

y dx xdy xdy y dx xdz zdx y dz zdy
or = 0
y 2 x x 2 y 2

x y z z
or d d d d = 0 ...(4)
y x x y
Integrating (4) we have

x y z z
= 0 (say) ...(5)
y x x y
or

x 2 y 2 ( z x y ) = cxy is the solution of equation (1).

Self Assessment

3. Solve the differential equation

2yzdx zxdy xy (1 z )dz

4. Solve the differential equation
2
xdx y dy a 2 x 2 y dz 0

Method II: Regarding one Variable as Constant

If the differential equation satisfies the condition of integrability and any two terms say
Pdx Qdy 0 can easily be integrated, then the third variable (say z) may be regarded as
constant so that dz 0.
Note that we should choose such a variable constant so that the remaining equation may be
integrated easily.
So the given differential equation will reduce to the integrable form

Pdx Qdy = 0 ...(1)

suppose its solution is
u = c (constant) ...(2)

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